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19.1: Thermodynamics and Heat Engines

Difficulty Level: At Grade Created by: CK-12

Key Concepts

  • The temperature of a gas is a measure of the amount of average kinetic energy that the atoms in the gas possess.
  • The pressure of a gas is the force the gas exerts on a certain area. For a gas in a container, the amount of pressure is directly related to the number and intensity of atomic collisions on a container wall.
  • An ideal gas is a gas for which interactions between molecules are negligible, and for which the gas atoms or molecules themselves store no potential energy. For an “ideal” gas, the pressure, temperature, and volume are simply related by the ideal gas law.
  • Atmospheric pressure (1 \;\mathrm{atm} = 101,000 Pascals) is the pressure we feel at sea level due to the weight of the atmosphere above us. As we rise in elevation, there is less of an atmosphere to push down on us and thus less pressure.
  • When gas pressure-forces are used to move an object then work is done on the object by the expanding gas. Work can be done on the gas in order to compress it.
  • Adiabatic process: a process that occurs with no heat gain or loss to the system in question.
  • Isothermal: a process that occurs at constant temperature (i.e. the temperature does not change during the process).
  • Isobaric: a process that occurs at constant pressure.
  • Isochoric: a process that occurs at constant volume.
  • If you plot pressure on the vertical axis and volume on the horizontal axis, the work done in any complete cycle is the area enclosed by the graph. For a partial process, work is the area underneath the curve, orP\Delta\!V.
  • In a practical heat engine, the change in internal energy must be zero over a complete cycle. Therefore, over a complete cycle W =  \Delta\!Q.
  • The work done by a gas during a portion of a cycle = P\Delta\!V, note \Delta\!Vcan be positive or negative.
  • The efficiency of any heat engine : \eta = W/Q_{in}
  • An ideal engine, the most efficient theoretically possible, is called a Carnot Engine. Its efficiency is given by the following formula, where the temperatures are, respectively, the temperature of the exhaust environment and the temperature of the heat input, in Kelvins. In a Carnot engine heat is input and exhausted in isothermal cycles, and the efficiency is  \eta = 1 -\frac{T_{\mathrm{cold}}}{T_\mathrm{hot}} .
  • The Stirling engine is a real life heat engine that has a cycle similar to the theoretical Carnot cycle. The Stirling engine is very efficient (especially when compared to a gasoline engine) and could become an important player in today's world where green energy and efficiency will reign supreme.

Key Equations

Q = mc \Delta T ; the heat gained or lost is equal to the mass of the object multiplied by its specific heat multiplied by the change of its temperature.

Q = mL ; the heat lost or gained by a substance due to a change in phase is equal to the mass of the substance multiplied by the latent heat of vaporization/fusion ( L refers to the latent heat)

1 cal = 4.184 Joules; your food calorie is actually a kilocalorie (Cal) and equal to 4184 J.

 Q_{\text{in}} = Q_{\text{out}} + W +\Delta U

U is the internal energy of the gas. (This is the first law of Thermodynamics and applies to all heat engines.)

 \left ({\frac{1}{2} m v ^2} \right )_{\text{avg}} = \frac{3}{2}kT

The average kinetic energy of atoms (each of mass m and average speed v) in a gas is related to the temperature T of the gas, measured in Kelvin. The Boltzmann constant k is a constant of nature, equal to 1.38\times10^{-23} \;\mathrm{J/K}.

 P = \frac{F}{A}

The pressure on an object is equal to the force pushing on the object divided by the area over which the force is exerted. Unit for pressure are \;\mathrm{N/m}^2 (called Pascals)

 PV = NkT

An ideal gas is a gas where the atoms are treated as point-particles and assumed to never collide or interact with each other. If you have N molecules of such a gas at temperature T and volume V, the pressure can be calculated from this formula. Note that k=1.38\times10^{-23}\;\mathrm{J/K}; this is the ideal gas law

 PV = nRT

V is the volume, n is the number of moles; R is the universal gas constant = 8.315 \;\mathrm{J/K-n}; this is the most useful form of the gas law for thermodynamics.

e_c  = 1-\frac{T_c}{T_h}  && \text{Efficiency of a Carnot (ideal) heat engine}

where  T_c and  T_h are the temperatures of the hot and cold reservoirs, respectively.

Molecular Kinetic Theory of a Monatomic Ideal Gas

The empirical combined gas law is simply a generalization of observed relationships. Using kinetic theory, it is possible to derive it from the principles of Newtonian mechanics. Previously, we thought of an ideal gas as one that obeys the combined gas law exactly. Within the current model, however, we can give a specific definition. We treat a monatomic ideal gas as a system of an extremely large number of very small particles in random motion that collide elastically between themselves and the walls of their container, where there are no interaction between particles other than collisions.

Consider some amount ( n atoms) of such a gas in a cubical container with side length  L . Let's trace the path of a single gas atom as it collides with the walls:

Ideal Gas Theory

The path of a single gas atom as it undergoes collisions with the walls of its container

Further, let's restrict ourselves to considering the motion of the particle along the x axis, and its collisions with the right y-z wall, as shown in the picture. Therefore, we only consider  v_x the component of the velocity vector perpendicular to the y-z wall.

If the particle's mass is m, in one collision, the particle's momentum in the  x direction changes by

 \Delta p = 2mv_x

Also, since it has to travel a distance  2 L (back and forth, basically) in the  x direction between collisions with the right  y-z wall, the time  \delta t between collisions will be

\Delta t = \frac{2L}{v_x}.

Ideal Gas Theory 2

Illustration x-direction atom from above.

According to Newton's second law, the force imparted by the single particle on the wall is

F = \frac{\Delta p}{\Delta t} = {2mv_x}\times \frac{v_x}{2L} = \frac{m{v_x}^2}{L}

Now, since there are  n (a very large number) atoms present, the net force imparted on the wall will be

F_{\text{net}} = n\times\frac{m({v_x}^2)_{\text{avg}}}{L}

Where the {v_x}^2 is averaged over all  n atoms.

Now let us attempt to relate this to the state variables we considered last chapter. Recall that pressure is defined as force per unit area:

 P = \frac{F}{A}

Since the area of the wall in question is  L^2 , the pressure exerted by the gas atoms on it will equal:

 P_{\text{net}} = \frac{F_{\text{net}}}{A} = n \frac{m({v_x}^2)_{\text{avg}}}{L\times L^2} = n \frac{m({v_x}^2)_{\text{avg}}}{L^3}

Since, for a cubical box, volume  V = L^3 , the formula above can be reduced to:

 P_{\text{net}} &= n \frac{m({v_x}^2)_{\text{avg}}}{V} \text{ or,} \\P_{\text{net}}V &= nm({v_x}^2)_{\text{avg}} &&\text{[1]}

By the Pythagorean theorem, any three-dimensional velocity vector has the following property:

 v^2 = {v_x}^2+{v_y}^2+{v_z}^2

Averaging this for the  n particles in the box, we get

 (v^2)_{\text{avg}} = ({v_x}^2)_{\text{avg}}+({v_y}^2)_{\text{avg}}+({v_z}^2)_{\text{avg}}

Since the motions of the particles are completely random (as stated in our assumptions), it follows that the averages of the squares of the velocity components should be equal: there is no reason the gas particles would prefer to travel in the  x direction over any other. In other words,

  ({v_x}^2)_{\text{avg}}=({v_y}^2)_{\text{avg}}=({v_z}^2)_{\text{avg}}

Plugging this into the average equation above, we find:

(v^2)_{\text{avg}} = 3\times ({v_x}^2)_{\text{avg}} &= 3\times ({v_y}^2)_{\text{avg}} = 3 \times ({v_z}^2)_{\text{avg}}\\\intertext{and} (v^2)_{\text{avg}}/3 &= ({v_x}^2)_{\text{avg}}

Plugging this into equation [1], we get:

P_{\text{net}} V = \frac{nm(v^2)_{\text{avg}}}{3} &&\text{[2]}

The left side of the equation should look familiar; this quantity is proportional to the average kinetic energy of the molecules in the gas, since

KE_{\text{avg}} = \frac{1}{2} m (v^2)_{\text{avg}}

Therefore, we have:

P_{\text{net}} V =\frac{2}{3} n (KE)_{\text{avg}} && \text{[3]}

This is a very important result in kinetic theory, since it expresses the product of two state variables, or system parameters, pressure and volume, in terms of an average over the microscopic constituents of the system. Recall the empirical ideal gas law from last chapter:

PV = nkT

The left side of this is identical to the left side of equation [3], whereas the only variable on the right side is temperature. By setting the left sides equal, we find:

\frac{2}{3} n (KE)_{\text{avg}} = nkT\intertext{or} T = \frac{2}{3k}(KE)_{\text{avg}}

Therefore, according to the kinetic theory of an monoatomic ideal gas, the quantity we called temperature is --- up to a constant coefficient --- a direct measure of the average kinetic energy of the atoms in the gas. This definition of temperature is much more specific and it is based essentially on Newtonian mechanics.

Temperature, Again

Now that we have defined temperature for a monoatomic gas, a relevant question is: can we extend this definition to other substances? It turns out that yes, we can, but with a significant caveat. In fact, according to classical kinetic theory, temperature is always proportional to the average kinetic energy of molecules in a substance. The constant of proportionality, however, is not always the same.

Consider: the only way to increase the kinetic energies of the atoms in a monoatomic gas is to increase their translational velocities. Accordingly, we assumed above that the kinetic energies of such atoms are stored equally in the three components ( x,y, \text{ and } z) of their velocities.

On the other hand, other gases --- called diatomic --- consist of two atoms held by a bond. This bond can be modeled as a spring, and the two atoms and bond together as a harmonic oscillator. Now, a single molecule's kinetic energy can be increased either by increasing its speed, by making it vibrate in simple harmonic motion, or by making it rotate around its center of mass. This difference is understood in physics through the concept of degrees of freedom: each degree of freedom for a molecule or particle corresponds to a possibility of increasing its kinetic energy independently of the kinetic energy in other degrees.

It might seem to you that monatomic gases should have one degree of freedom: their velocity. They have three because their velocity can be altered in one of three mutually perpendicular directions without changing the kinetic energy in other two --- just like a centripetal force does not change the kinetic energy of an object, since it is always perpendicular to its velocity. These are called translational degrees of freedom.

Diatomic gas molecules, on the other hand have more: the three translational explained above still exist, but there are now also vibrational and rotational degrees of freedom. Monatomic and diatomic degrees of freedom can be illustrated like this:

Temperature is an average of kinetic energy over degrees of freedom, not a sum. Let's try to understand why this is in reference to our monoatomic ideal gas. In the derivation above, volume was constant; so, temperature was essentially proportional to pressure, which in turn was proportional to the kinetic energy due to translational motion of the molecules. If the molecules had been able to rotate as well as move around the box, they could have had the same kinetic energy with slower translational velocities, and, therefore, lower temperature. In other words, in that case, or assumption that the kinetic energy of the atoms only depends on their velocities, implied between equations [2] and [3], would not have held. Therefore, the number of degrees of freedom in a substance determines the proportionality between molecular kinetic energy and temperature: the more degrees of freedom, the more difficult it will be to raise its temperature with a given energy input.

In solids, degrees of freedom are usually entirely vibrational; in liquids, the situation becomes more complicated. We will not attempt to derive results about these substances, however.

A note about the above discussion: since the objects at the basis of our understanding of thermodynamics are atoms and molecules, quantum effects can make certain degrees of freedom inaccessible at specific temperature ranges. Unlike most cases in your current physics class, where these can be ignored, in this case, quantum effects can make an appreciable difference. For instance, the vibrational degrees of freedom of diatomic gas molecules discussed above are, for many gases, inaccessible in very common conditions, although we do not have the means to explain this within our theory. In fact, this was one of the first major failures of classical physics that ushered in the revolutionary discoveries of the early 20th century.

Thermal Energy

In light of the above derivation, it should not surprise you that the kinetic energy from motion of molecules contributes to what is called the thermal energy of a substance. This type of energy is called sensible energy. In ideal gases, this is the only kind of thermal energy present.

Solids and liquids also have a different type of thermal energy as well, called Latent Energy, which is associated with potential energy of their intermolecular bonds in that specific phase --- for example the energy it takes to break the bonds between water molecules in melting ice (remember, we assumed molecules do not interact in the ideal gas approximation).

To recap, there are two types of Thermal Energy:

  • The kinetic energy from the random motion of the molecules or atoms of the substance, called Sensible Energy
  • The intermolecular potential energy associated with changes in the phase of a system (called Latent Energy).

Heat

The term heat is formally defined as a transfer of thermal energy between substances. Note that heat is not the same as thermal energy. Before the concept of thermal energy, physicists sometimes referred to the 'heat energy' of a substance, that is, the energy it received from actual 'heating' (heating here can be understood as it is defined above, though for these early physicists and chemists it was a more 'common sense' idea of heating: think beaker over Bunsen burner). The idea was then to try to explain thermodynamic phenomena through this concept.

The reason this approach fails is that --- as stated in the paragraphs above --- it is in fact thermal energy that is most fundamental to the science, and 'heating' is not the only way to change the thermal energy of a substance. For example, if you rub your palms together, you increase the thermal energy of both palms.

Once heat (a transfer of thermal energy) is absorbed by a substance, it becomes indistinguishable from the thermal energy already present: what methods achieved that level of thermal energy is no longer relevant. In other words, 'to heat' is a well defined concept as a verb: its use automatically implies some kind of transfer. When heat using as a noun, one needs to be realize that it must refer to this transfer also, not something that can exist independently.

Specific Heat Capacity and Specific Latent Heat

The ideas in the paragraphs above can be understood better through the concept of specific heat capacity (or specific heat for short), which relates an increase in temperature of some mass of a substance to the amount of heat required to achieve it. In other words, for any substance, it relates thermal energy transfers to changes in temperature. It has units of Joules per kilogram Kelvin. Here is how we can define and apply specific heat ( Q refers to heat supplied,  m to the mass of the substance and  c to its specific heat capacity):

Q = cm\Delta T && \text{[2]}

Heat capacity is largely determined by the number of degrees of freedom of the molecules in a substance (why?). However, it also depends on other parameters, such as pressure. Therefore, the formula above implicitly assumes that these external parameters are held constant (otherwise we wouldn't know if we're measuring a change in specific heat is real or due to a change in pressure).

When a substance undergoes a phase change, its temperature does not change as it absorbs heat. We referred to this as an increase or decrease in latent energy earlier. In this case, the relevant question is how much heat energy does it require to change a unit mass of the substance from one phase to another? This ratio is known as latent heat, and is related to heat by the following equation ( L refers to the latent heat):

Q = Lm  && \text{[3]}

During a phase change, the number of degrees of freedom changes, and so does the specific heat capacity. Heat capacity can also depend on temperature within a given phase, but many substances, under constant pressure, exhibit a constant specific heat over a wide range of temperatures. For instance, look at the graph of temperature vs heat input for a mole ( 6.0221415 \times 10^{23} molecules) of water at http://en.wikipedia.org/wiki/File:Energy_thru_phase_changes.png. Note that the x-axis of the graph is called 'relative heat energy' because it takes a mole of water at 0 degrees Celsius as the reference point.

The sloped segments on the graph represent increases in temperature. These are governed by equation [1]. The flat segments represent phase transitions, governed by equation [2]. Notice that the sloped segments have constant, though different, slopes. According to equation [1], the heat capacity at any particular phase would be the slope of the segment that corresponds to that phase on the graph. The fact that the slopes are constant means that, within a particular phase, the heat capacity does not change significantly as a function of temperature.

Table of Specific Heat Values
Substance Specific Heat, c \;(cal/g^\circ C)
Air 6.96
Water 1.00
Alcohol 0.580
Steam 0.497
Ice (-10^\circ C) 0.490
Aluminum 0.215
Zinc 0.0925
Brass 0.0907
Silver 0.0558
Lead 0.0306
Gold \sim Lead 0.0301
Table of Heat of Vaporization
Substance Fusion, L_f (cal/g) Vaporization, L_v (cal/g)
Water 80.0 540
Alcohol 26 210
Silver 25 556
Zinc 24 423
Gold 15 407
Helium - 5.0

Entropy

The last major concept we are going to introduce in this chapter is entropy. We noted earlier that temperature is determined not just by how much thermal energy is present in a substance, but also how it can be distributed. Substances whose molecules have more degrees of freedom will generally require more thermal energy for an equal temperature increase than those whose molecules have fewer degrees of freedom.

Entropy is very much related to this idea: it quantifies how the energy actually is distributed among the degrees of freedom available. In other words, it is a measure of disorder in a system. An example may illustrate this point. Consider a monatomic gas with  N atoms (for any appreciable amount of gas, this number will be astronomical). It has  3N degrees of freedom. For any given value of thermal energy, there is a plethora of ways to distribute the energy among these. At one extreme, it could all be concentrated in the kinetic energy of a single atom. On the other, it could be distributed among them all. According to the discussion so far, these systems would have the same temperature and thermal energy. Clearly, they are not identical, however. This difference is quantified by entropy: the more evenly distributed the energy, the higher the entropy of the system. Here is an illustration:

The Laws of Thermodynamics

Now that we have defined the terms that are important for an understanding of thermodynamics, we can state the laws that govern relevant behavior. These laws, unlike Newton's Laws or Gravity, are not based on new empirical observations: they can be derived based on statistics and known principles, such as conservation of energy. By understanding the laws of thermodynamics we can analyze heat engines, or machines that use heat energy to perform mechanical work.

The First Law

The First Law of Thermodynamics is simply a statement of energy conservation applied to thermodynamics systems: the change in the internal --- for our purposes, this is the same as thermal --- energy (denoted   \Delta U ) of a closed system is equal to the difference of net input heat and performed work. In other words,

\Delta  U = Q_{net} -  W  && \text{[4] First Law}

Note that this does not explain how the system will transform input heat to work, it simply enforces the energy balance.

The Second Law

The Second Law of Thermodynamics states that the entropy of an isolated system will always increase until it reaches some maximum value. Consider it in light of the simplified example in the entropy section: if we allow the low entropy system to evolve, it seems intuitive collisions will eventually somehow distribute the kinetic energy among the atoms.

The Second Law generalizes this intuition to all closed thermodynamic systems. It is based on the idea that in a closed system, energy will be randomly exchanged among constituent particles --- like in the simple example above --- until the distribution reaches some equilibrium (again, in any macroscopic system there will be an enormous number of atoms, degrees of freedom, etc). Since energy is conserved in closed systems, this equilibrium has to preserve the original energy total. In this equilibrium, the Second Law --- fundamentally a probabilistic statement --- posits that the energy will be distributed in the most likely way possible. This typically means that energy will be distributed evenly across degrees of freedom.

This allows us to formulate the Second Law in another manner, specifically: heat will flow spontaneously from a high temperature region to a low temperature region, but not the other way. This is just applying the thermodynamic vocabulary to the logic of the above paragraph: in fact, this is the reason for the given definition of temperature. When two substances are put in thermal contact (that is, they can exchange thermal energy), heat will flow from the system at the higher temperature (because it has more energy in its degrees of freedom) to the system with lower temperature until their temperatures are the same.

When a single system is out equilibrium, there will be a net transfer of energy from one part of it to another. In equilibrium, energy is still exchanged among the atoms or molecules, but not on a system-wide scale. Therefore, entropy places a limit on how much work a system can perform: the higher the entropy, the more even the distribution of energy, the less energy available for transfer.

Heat Engines

Heat engines transform input heat into work in accordance with the laws of thermodynamics. For instance, as we learned in the previous chapter, increasing the temperature of a gas at constant volume will increase its pressure. This pressure can be transformed into a force that moves a piston.

The mechanics of various heat engines differ but their fundamentals are quite similar and involve the following steps:

  1. Heat is supplied to the engine from some source at a higher temperature  (T_h) .
  2. Some of this heat is transferred into mechanical energy through work done  (W) .
  3. The rest of the input heat is transferred to some source at a lower temperature  (T_c) until the system is in its original state.

A single cycle of such an engine can be illustrated as follows:

In effect, such an engine allows us to 'siphon off' part of the heat flow between the heat source and the heat sink. The efficiency of such an engine is define as the ratio of net work performed to input heat; this is the fraction of heat energy converted to mechanical energy by the engine:

 e =  \frac{W}{Q_i} && \text{[5] Efficiency of a heat engine}

If the engine does not lose energy to its surroundings (of course, all real engines do), then this efficiency can be rewritten as

 e = \frac{Q_i-Q_o}{Q_i}  && \text{[6] Efficiency of a lossless heat engine}

A Carnot Engine, the most efficient heat engine possible, has an efficiency equal to

e_c  = 1-\frac{T_c}{T_h}  && \text{[7] Efficiency of a Carnot (ideal) heat engine}

where  T_c and  T_h are the temperatures of the hot and cold reservoirs, respectively.

Some Important Points

  • In a practical heat engine, the change in internal energy must be zero over a complete cycle. Therefore, over a complete cycle W =  \Delta Q.
  • The work done by a gas during a portion of a cycle = P\Delta\!V, note \Delta\!Vcan be positive or negative.

Gas Heat Engines

  • When gas pressure-forces are used to move an object then work is done on the object by the expanding gas. Work can be done on the gas in order to compress it.
  • If you plot pressure on the vertical axis and volume on the horizontal axis (see  P-V diagrams in the last chapter), the work done in any complete cycle is the area enclosed by the graph. For a partial process, work is the area underneath the curve, orP\Delta\!V.

Question: A heat engine operates at a temperature of 650\mathrm{K}. The work output is used to drive a pile driver, which is a machine that picks things up and drops them. Heat is then exhausted into the atmosphere, which has a temperature of 300\mathrm{K}.

a) What is the ideal efficiency of this engine?

b.) The engine drives a 1200\mathrm{kg} weight by lifting it 50m in 2.5 sec. What is the engine’s power output?

c) If the engine is operating at 50\% of ideal efficiency, how much power is being consumed?

d) The fuel the engine uses is rated at 2.7\times  10^6\mathrm{J/kg}. How many kg of fuel are used in one hour?

Answer:

a) We will plug the known values into the formula to get the ideal efficiency.

\eta = 1 -\frac{T_{\mathrm{cold}}}{T_\mathrm{hot}}=1-\frac{300\mathrm{K}}{650\mathrm{K}}=54\%

b) To find the power of the engine, we will use the power equation and plug in the known values.

P=\frac{W}{t}=\frac{Fd}{t}=\frac{mad}{t}=\frac{1200\mathrm{kg}\times  9.8\mathrm{m/s^2}\times 50\mathrm{m}}{2.5\mathrm{sec}}=240\mathrm{kW}

c) First, we know that it is operating at 50\% of ideal efficiency. We also know that the max efficiency of this engine is 54\%. So the engine is actually operating at

.5\times  54\%=27\%

of 100\% efficiency. So 240\mathrm{kW} is 27\% of what?

.27x=240\mathrm{kW}  \Rightarrow x=\frac{240\mathrm{kW}}{.27}=890\mathrm{kW}

Thermodynamics and Heat Engines Problem Set

  1. Consider a molecule in a closed box. If the molecule collides with the side of the box, how is the force exerted by the molecule on the box related to the momentum of the molecule? Explain conceptually, in words rather than with equations.
  2. If the number of molecules is increased, how is the pressure on a particular area of the box affected? Explain conceptually, in words rather than with equations.
  3. The temperature of the box is related to the average speed of the molecules. Use momentum principles to relate temperature to pressure. Explain conceptually, in words rather than with equations.
  4. What would happen to the number of collisions if temperature and the number of molecules remained fixed, but the volume of the box increased? Explain conceptually, in words rather than with equations.
  5. Use the reasoning in the previous four questions to qualitatively derive the ideal gas law.
  6. Typical room temperature is about 300 \;\mathrm{K}. As you know, the air in the room contains both O_2 and N_2 gases, with nitrogen the lower mass of the two. If the average kinetic energies of the oxygen and nitrogen gases are the same (since they are at the same temperature), which gas has a higher average speed?
  7. Use the formula P = F / A to argue why it is easier to pop a balloon with a needle than with a finger (pretend you don’t have long fingernails).
  8. Take an empty plastic water bottle and suck all the air out of it with your mouth. The bottle crumples. Why, exactly, does it do this?
  9. You will notice that if you buy a large drink in a plastic cup, there will often be a small hole in the top of the cup, in addition to the hole that your straw fits through. Why is this small hole necessary for drinking?
  10. Suppose you were swimming in a lake of liquid water on a planet with a lower gravitational constant g than Earth. Would the pressure 10 meters under the surface be the same, higher, or lower, than for the equivalent depth under water on Earth? (You may assume that the density of the water is the same as for Earth.)
  11. Why is it a good idea for Noreen to open her bag of chips before she drives to the top of a high mountain?
  12. Explain, using basic physics conservation laws, why the following conditions would cause the ideal gas law to be violated:
    1. There are strong intermolecular forces in the gas.
    2. The collisions between molecules in the gas are inelastic.
    3. The molecules are not spherical and can spin about their axes.
    4. The molecules have non-zero volume.

    To the right is a graph of the pressure and volume of a gas in a container that has an adjustable volume. The lid of the container can be raised or lowered, and various manipulations of the container change the properties of the gas within. The points a, b, and c represent different stages of the gas as the container undergoes changes (for instance, the lid is raised or lowered, heat is added or taken away, etc.) The arrows represent the flow of time. Use the graph to answer the following questions.

  13. Consider the change the gas undergoes as it transitions from point b to point c. What type of process is this?
    1. adiabatic
    2. isothermal
    3. isobaric
    4. isochoric
    5. entropic
  14. Consider the change the gas undergoes as it transitions from point c to point a. What type of process is this?
    1. adiabatic
    2. isothermal
    3. isobaric
    4. isochoric
    5. none of the above
  15. Consider the change the gas undergoes as it transitions from point a to point b. Which of the following best describes the type of process shown?
    1. isothermal
    2. isobaric
    3. isochoric
  16. How would an isothermal process be graphed on a P-V diagram?
  17. Write a scenario for what you would do to the container to make the gas within undergo the cycle described above.
  18. Why is it so cold when you get out of the shower wet, but not as cold if you dry off first before getting out of the shower? _____________________________________________________________
  19. Antonio is heating water on the stove to boil eggs for a picnic. How much heat is required to raise the temperature of his 10.0-kg vat of water from 20^\circ C to 100^\circ C?
  20. Amy wishes to measure the specific heat capacity of a piece of metal. She places the 75-g piece of metal in a pan of boiling water, then drops it into a styrofoam cup holding 50 g of water at 22^\circ C. The metal and water come to an equilibrium temperature of 25^\circ C. Calculate:
    1. The heat gained by the water
    2. The heat lost by the metal
    3. The specific heat of the metal
  21. John wishes to heat a cup of water to make some ramen for lunch. His insulated cup holds 200 g of water at 20^\circ C. He has an immersion heater rated at 1000 W (1000 J/s) to heat the water.
    1. How many JOULES of heat are required to heat the water to 100^\circ C?
    2. How long will it take to do this with a 1000-W heater?
    3. Convert your answer in part b to minutes.
  22. You put a 20g cylinder of aluminum (c=0.2 \ cal/g/^\circ C) in the freezer (T=-10^\circ C). You then drop the aluminum cylinder into a cup of water at 20^\circ C. After some time they come to a common temperature of 12^\circ C. How much water was in the cup?
  23. Emily is testing her baby’s bath water and finds that it is too cold, so she adds some hot water from a kettle on the stove. If Emily adds 2.00 kg of water at 80.0^\circ C to 20.0 kg of bath water at 27.0^\circ C, what is the final temperature of the bath water?
  24. You are trying to find the specific heat of a metal. You heated a metal in an oven to 250^\circ C. Then you dropped the hot metal immediately into a cup of cold water. To the right is a graph of the temperature of the water versus time that you took in the lab. The mass of the metal is 10g and the mass of the water is 100g. Recall that water has a specific heat of 1 \ cal/g^\circ C.
  25. How much heat is required to melt a 20 g cube of ice if
    1. the ice cube is initially at 0^\circ C
    2. the ice cube is initially at -20^\circ C (be sure to use the specific heat of ice)
  26. A certain alcohol has a specific heat of 0.57 \ cal/g^\circ C and a melting point of -114^\circ C. You have a 150 g cup of liquid alcohol at 22^\circ C and then you drop a 10 g frozen piece of alcohol at -114^\circ C into it. After some time the alcohol cube has melted and the cup has come to a common temperature of 7^\circ C. a. What is the latent heat of fusion (i.e. the ‘L’ in the Q = mL equation) for this alcohol? b. Make a sketch of the graph of the alcohol’s temperature vs. time c. Make a sketch of the graph of the water’s temperature vs. time
  27. Calculate the average speed of N_2 molecules at room temperature (300 \;\mathrm{K}). (You remember from your chemistry class how to calculate the mass (in kg) of an N_2 molecule, right?)
  28. How high would the temperature of a sample of O_2 gas molecules have to be so that the average speed of the molecules would be 10% the speed of light?
  29. How much pressure are you exerting on the floor when you stand on one foot? (You will need to estimate the area of your foot in square meters.)
  30. Calculate the amount of force exerted on a 2 \;\mathrm{cm} \times 2 \;\mathrm{cm} patch of your skin due to atmospheric pressure (P_0 = 101,000 \;\mathrm{Pa}). Why doesn’t your skin burst under this force?
  31. Use the ideal gas law to estimate the number of gas molecules that fit in a typical classroom.
  32. Assuming that the pressure of the atmosphere decreases exponentially as you rise in elevation according to the formula P = P_0 e^\frac{-h} {a}, where P_0 is the atmospheric pressure at sea level (101,000 \;\mathrm{Pa}), h is the altitude in km and a is the scale height of the atmosphere (a \approx 8.4 \;\mathrm{km}).
    1. Use this formula to determine the change in pressure as you go from San Francisco to Lake Tahoe, which is at an elevation approximately 2 \;\mathrm{km} above sea level.
    2. If you rise to half the scale height of Earth’s atmosphere, by how much does the pressure decrease?
    3. If the pressure is half as much as on sea level, what is your elevation?
  33. Ata university the following experiment was conducted by a professor. A rock was dropped from the roof of the lab and, with expensive equipment, was observed to gain 100 \;\mathrm{J} of internal energy. The professor explained to his students that the law of conservation of energy required that if he put 100 \;\mathrm{J} of heat into the rock, the rock would then rise to the top of the building. When this did not occur, the professor declared the law of conservation of energy invalid.
    1. Was the law of conservation of energy violated in this experiment, as was suggested? Explain.
    2. If the law wasn’t violated, then why didn’t the rock rise?
  34. An instructor has an ideal monatomic helium gas sample in a closed container with a volume of 0.01\;\mathrm{m}^3, a temperature of 412 \;\mathrm{K}, and a pressure of 474 \;\mathrm{kPa}.
    1. Approximately how many gas atoms are there in the container?
    2. Calculate the mass of the individual gas atoms.
    3. Calculate the speed of a typical gas atom in the container.
    4. The container is heated to 647 \;\mathrm{K}. What is the new gas pressure?
    5. While keeping the sample at constant temperature, enough gas is allowed to escape to decrease the pressure by half. How many gas atoms are there now?
    6. Is this number half the number from part (a)? Why or why not?
    7. The closed container is now compressed isothermally so that the pressure rises to its original pressure. What is the new volume of the container?
    8. Sketch this process on a P-V diagram.
    9. Sketch cubes with volumes corresponding to the old and new volumes.
  35. A famous and picturesque dam, 80 \;\mathrm{m} high, releases 24,000 \;\mathrm{kg} of water a second. The water turns a turbine that generates electricity.
    1. What is the dam’s maximum power output? Assume that all the gravitational potential energy of the water is converted into electrical energy.
    2. If the turbine only operates at 30% efficiency, what is the power output?
    3. How many Joules of heat are exhausted into the atmosphere due to the plant’s inefficiency?

  36. A heat engine operates at a temperature of 650 \;\mathrm{K}. The work output is used to drive a pile driver, which is a machine that picks things up and drops them. Heat is then exhausted into the atmosphere, which has a temperature of 300 \;\mathrm{K}.
    1. What is the ideal efficiency of this engine?
    2. The engine drives a 1200 \;\mathrm{kg} weight by lifting it 50\;\mathrm{m} in 2.5 \;\mathrm{sec}. What is the engine’s power output?
    3. If the engine is operating at 50% of ideal efficiency, how much power is being consumed?
    4. How much power is exhausted?
    5. The fuel the engine uses is rated at 2.7\times10^6 \;\mathrm{J/kg}. How many kg of fuel are used in one hour?
  37. Calculate the ideal efficiencies of the following sci-fi heat engines:
    1. A nuclear power plant on the moon. The ambient temperature on the moon is 15 \;\mathrm{K}. Heat input from radioactive decay heats the working steam to a temperature of 975 \;\mathrm{K}.
    2. A heat exchanger in a secret underground lake. The exchanger operates between the bottom of a lake, where the temperature is 4 \;\mathrm{C}, and the top, where the temperature is 13 \;\mathrm{C}.
    3. A refrigerator in your dorm room at Mars University. The interior temperature is 282 \;\mathrm{K}; the back of the fridge heats up to 320 \;\mathrm{K}.
  38. How much external work can be done by a gas when it expands from 0.003 \;\mathrm{m}^3 to 0.04 \;\mathrm{m}^3 in volume under a constant pressure of 400 \;\mathrm{kPa}? Can you give a practical example of such work?
  39. In the above problem, recalculate the work done if the pressure linearly decreases from 400 \;\mathrm{kPa} to 250\;\mathrm{kPa} under the same expansion. Hint: use a PV diagram and find the area under the line.
  40. One mole (N = 6.02\times10^{23}) of an ideal gas is moved through the following states as part of a heat engine. The engine moves from state A to state B to state C, and then back again. Use the Table (below) to answer the following questions:
    1. Draw a P-V diagram.
    2. Determine the temperatures in states A, B, and C and then fill out the table.
    3. Determine the type of process the system undergoes when transitioning from A to B and from B to C. (That is, decide for each if it is isobaric, isochoric, isothermal, or adiabatic.)
    4. During which transitions, if any, is the gas doing work on the outside world? During which transitions, if any, is work being done on the gas?
    5. What is the amount of net work being done by this gas?
State Volume (m3) Pressure (atm) Temperature (K)
A 0.01 0.60
B 0.01 0.25
C 0.02 0.25
  1. A sample of gas is used to drive a piston and do work. Here’s how it works:
    • The gas starts out at standard atmospheric pressure and temperature. The lid of the gas container is locked by a pin.
    • The gas pressure is increased isochorically through a spigot to twice that of atmospheric pressure.
    • The locking pin is removed and the gas is allowed to expand isobarically to twice its volume, lifting up a weight. The spigot continues to add gas to the cylinder during this process to keep the pressure constant.
    • Once the expansion has finished, the spigot is released, the high-pressure gas is allowed to escape, and the sample settles back to 1 \;\mathrm{atm}.
    • Finally, the lid of the container is pushed back down. As the volume decreases, gas is allowed to escape through the spigot, maintaining a pressure of 1 \;\mathrm{atm}. At the end, the pin is locked again and the process restarts.
    1. Draw the above steps on a P-V diagram.
    2. Calculate the highest and lowest temperatures of the gas.
  2. A heat engine operates through 4 cycles according to the PV diagram sketched below. Starting at the top left vertex they are labeled clockwise as follows: a, b, c, and d.
    1. From a-b the work is 75 \;\mathrm{J} and the change in internal energy is 100 \;\mathrm{J}; find the net heat.
    2. From the a-c the change in internal energy is -20 \;\mathrm{J}. Find the net heat from b-c.
    3. From c-d the work is -40 \;\mathrm{J}. Find the net heat from c-d-a.
    4. Find the net work over the complete 4 cycles.
    5. The change in internal energy from b-c-d is -180 \;\mathrm{J}. Find:
      1. the net heat from c-d
      2. the change in internal energy from d-a
      3. the net heat from d-a

  3. A 0.1 sample mole of an ideal gas is taken from state A by an isochoric process to state B then to state C by an isobaric process. It goes from state C to D by a process that is linear on a PV diagram, and then it goes back to state A by an isobaric process. The volumes and pressures of the states are given in the Table (below); use this data to complete the following:
    1. Find the temperature of the 4 states
    2. Draw a PV diagram of the process
    3. Find the work done in each of the four processes
    4. Find the net work of the engine through a complete cycle
    5. If 75 \;\mathrm{J} of heat is exhausted in D-A and A-B and C-D are adiabatic, how much heat is inputted in B-C?
    6. What is the efficiency of the engine?
state Volume in m^3 \times 10^{-3} Pressure in N/m^2 \times 10^5
A 1.04 2.50
B 1.04 4.00
C 1.25 4.00
D 1.50 2.50

Answers to Selected Problems

  1. In each collision, the perpendicular (to the box side) component of the molecule's momentum has to be reversed. Therefore, a higher momentum molecule will, on average, exert a larger force on the box.
  2. Intuitively, we can say that there will be more collisions; therefore, the pressure will increase.
  3. Faster molecules imply larger, on average, perpendicular components of momentum in collisions and therefore higher pressure on the sides of the box. So higher temperature would be related to higher pressure.
  4. As there would be "more room" for the molecules to occupy, the number of collisions per unit of time would decrease. To see this more clearly we can consider reducing the volume of the box; this would eventually lead to more collisions per time as space runs out (this is not due to the molecules themselves, which have zero volume by assumption, but due to the longest collision-free distance possible being equal to the diagonal of such a box). We would expect the opposite trend going in the other direction.
  5. The previous questions imply that pressure is directly related to the number of molecules (2) and temperature (3), and inversely to volume (4), suggesting a law of the form  P = \frac{kNT}{V} , where k is some constant.
  6. Assuming the average kinetic energies are equal, we can write \frac{1}{2}m_ov_o^2 = \frac{1}{2}m_nv_n^2 \implies v_o = v_n\sqrt{\frac{m_n}{m_o}}. Since the ratio of masses is less than 1, oxygen will have a lower average speed.
  7. To pop a balloon a threshold pressure must be achieved. Since the surface area of a needle is smaller than that of your finger, it will exert a higher pressure with the same amount of force. The size of the hole is not relevant since a balloon is popped by a hole of any size.
  8. .The bottle crumples since it can no longer push back against atmospheric pressure. Similar effects can be observed when you close a bottle on top of a mountain and drive down, or due to changes in weather (in these cases the outside, rather than inside, pressure changes).
  9. Drinking some of the liquid increases the empty volume in the cup, and so would decrease pressure on its walls if the number of gas molecules inside were constant (see problem 4 or empirical gas laws). Informally, we can see that the pressure would have to be increased somehow if we didn't want the cup to crumple due to air pressure. By the ideal gas law, we can see that the volume would decrease at constant temperature (the cup would crumple) unless more molecules are let in. People chugging bottled water after intense exercise should have had this experience, since typical plastic water bottles don't have such an inlet.
  10. Since water (as well as air) pressure is basically due to the weight of the water or air above you, the pressure would be lower on the planet with a lower  g, where this water would weigh less. It would still have the same mass, though.
  11. On top of a mountain, the air pressure is lower than at sea level since there is less air pushing down on you (some of the atmosphere is below you), and so the bag of chips would be more inflated there. This would make it more difficult to open without spilling.
  12. Part (a) would violate the non-interaction assumption. If the collisions are inelastic (part b), the gas would continuously lose energy to the environment, violating the reversibility assumption. If the molecules can spin about their axes (part c), they could have rotational kinetic energy, which violates the assumption that the only source of energy for the gas molecules is translational kinetic. Finally, part (d) would imply that the number of molecules would limit available volume, which also violates assumptions used to derive the ideal gas law.
  13. Volume stays constant through this process, so it is isochoric.
  14. Here pressure stays constant, so it is isobaric.
  15. This process is isothermal since neither pressure nor volume is held constant.
  16. Using the formula  PV = NRT \implies P = \frac{NRT}{V} . Holding the numerator constant (as is true for isothermal processes), we could plug in different values for volume and "connect the dots". This will result in an inverse relationship between pressure and volume.
  17. An increase in pressure at constant volume can be achieved by heating the container while it is kept from expanding. It could be put into a water bath at a higher temperature. A decrease in volume at constant pressure can be achieved by having one side of the container act like a piston and exposing it to air pressure or some other source of constant pressure. If the volume decreases at constant pressure, the gas must cool. Then allowing the piston to expand while the container is in a (relatively large) water bath at its temperature would approximate an isothermal process.
  18. .
  19. 800,000 cal or 3360 kJ
    1. 150 cal (630 J)
    2. same as a!
    3. 0.027 \ cal/g^\circ C \ (0.11 \ J/g^\circ C)
    1. 67,000 J
    2. 67.2 s
    3. 1.1 min
  20. 11.0 g
  21. 31.8^\circ C
  22. 0.44 \ cal/g^\circ C
    1. 1600 cal (6720 J)
    2. 1800 cal (7560 J)
  23. 59.3 cal/g
  24. We can use the equation   {\frac{1}{2} m {v^2}_{\text{avg}}} = \frac{3}{2}kT to get  v_\text{avg} = \sqrt\frac{3kT}{m} . Then using the values  m \approx 4.65\times 10^{-26}\text{kg},k= 1.38\times10^{-23}  \;\mathrm{J/K} and T = 300, we get  v_\text{avg} \approx 517\text{m/s}.
  25. We can again use the equation   {\frac{1}{2} m {v^2}_{\text{avg}}} = \frac{3}{2}kT, but this time we would like  v_\text{avg} = .1\times 3\times 10^8\text{m/s}. Using m \approx 5.36 \times 10^{-26} \text{kg} and k= 1.38\times10^{-23}  \;\mathrm{J/K} in  T = \frac{mv^2_{avg}}{3k} , we get  T \approx 1.17\times 10^{12}\text{K}.
  26. Assuming my foot is a rectangle 5cm wide and 30cm long and my weight is 50kg, we can calculate the pressure as the force (due to gravity) divided by area. Converting to meters, we get P=\frac{F}{A} =\frac{50 \times 9.8}{5\times 10^{-2} \times 30 \times10^{-2}}\approx 1.67 \times 10^{5}\text{Pa}.
  27. #We can use  P=\frac{F}{A}, this time with the pressure and force known. So,  1.01\times 10^5 = \frac{F}{4\times10^{-4}}, so  F= 1.01\times 10^5 \times 4\times10^{-4}\approx 40\text{N}. Your skin does not burst because there's an equal amount of pressure outward; in other words, there is no net force on your skin.
  28. We can assume the room is a rectangular prism, or shoebox, shape. Reasonable measures for length, width, and height could be 10 meters, 9 meters, and 5 meters. In this case the volume would be given by their product.; that is,  V = 450 \text{m}^3. Like in the previous problems, we can use  T = 300\text{K} and  k= 1.38\times10^{-23}  \;\mathrm{J/K}. Assuming the classroom is at sea level, we can use  P = 10^5 \text{Pa}. Rearranging the ideal gas law, we get the formula  N =\frac{PV}{kT} =\frac{10^5\times450}{1.38\times10^{-23}\times300} \approx 10^{28} \text{ molecules}.
  29. Plugging the values given into the formula, we get pressure as a function of height:  P(h) =10^5\times e^{-h/8400}
    1. To calculate the change in pressure we can subtract the pressure at 2 kilometers from that at sea level:  \Delta P  = P(2000)-p(0) = 10^5\times e^{-2000/8400}-10^5\approx -2.1\times 10^5 \text{Pa}. The change is negative since the pressure decreases as your height relative to the atmosphere increases.
    2. Analogously,  \Delta P = P(4200)-p(0) = 10^5\times e^{-1/2}-10^5\approx -3.9 \times 10^5 \text{Pa}, or\ a\  3.9 \times 10^5 Pa decrease.
    3. We can use the pressure as a function of height formula from above by setting it equal to half the atmospheric pressure. The atmospheric pressure cancels from both sides:  P(h) = \cancel{10^5}\times e^{-h/8400} = .5\times \cancel{10^5} , and we are left with  e^{-h/8400} = .5 . Taking the natural logarithm of both sides gives  \text{ln}(e^{-h/8400})=\text{ln}(.5) \implies -h/8400 =\text{ln}(.5)\implies h = -8400\times \text{ln}(.5) \approx 5.8\text{km}
    1. Conservation of energy was not violated, because it does not state that heat must be converted into gravitational potential energy as the problem seems to imply. It simply states that the total energy in any closed system must remain constant.
    2. See part a.
    1. As  PV =NkT \implies N = PV/(kT) = 474\times 10^3(.01)/(412\times1.38\times10^{-23}) , which is about  8.34 \times 10^{23} atoms.
    2. The mass of a Helium atom is 6.69 \times 10^{-27}kg.
    3. Using the formula 1/2mv_{\text{ave}}^2=3/2kT \implies v = \sqrt{3kT/m} = \sqrt{\frac{3\times1.38\times10^{-23}\times 412}{6.69\times10^{-27}}} \approx 1.6\times 10^3m/s.
    4. If volume does not change, we can use the formula from part a (the Ideal Gas Law) to find the new pressure. Specifically,  P_\text{new} = NkT_\text{new}/V = P_\text{old}\times T_\text{new}/T_\text{old} = 474\times 647/412 \approx 744 \text{kPa} .
    5. Again, since only pressure is changing on the left side of Ideal Gas Law and only the number of atoms on the right side,  N_\text{new} = .5P_\text{new}V/kT\text{new} =.5P_\text{old}V/kT_\text{old}=.5N_\text{old} . Since the number of atoms didnt change in the previous heating, the right side will equal half the original number, or about  4.2 \times 10^{23} atoms.
    6. Yes, since the number of atoms didn't change in the heating.
    7. We can make use of the fact that in the equation for an ideal gas, pressure and volume are inversely proportional if the number of atoms is constant at constant temperature. Therefore, since the pressure is now 744/2=372 kPa and we would like it to be 474 kPa, the volume has to decrease by a factor of  474/372, meaning  V_\text{new} = .01\times(474/372)^{-1} \approx .00785\text{m}^3.
  30. Using the formula  U_g = mg\Delta h , we calculate that the water gains 24,000 \times 9.8 \times 80 \approx 1.9\times 10^7 \text{J} of kinetic energy per second.
    1. Therefore the maximum output would be  1.9\times 10^7 Watts.
    2. Now the output would be .3\times 1.9\times 10^7 \approx 5.6 \times 10^6 Watts.
    3. The difference is between a and b, or  1.3 \times 10^7 \text{W} is exhausted into the atmosphere.
  31. Done in chapter as example problem.
    1. 54%
    2. 240 \;\mathrm{kW}
    3. 890 \;\mathrm{kW}
    4. 590 \;\mathrm{kW}
    5. 630 \;\mathrm{kg}
  32. Using the formula  e_\text{ideal} = 1-T_c/T_h :
    1.  e_\text{ideal} = 1- 15/975 \approx 98%
    2. First we convert to Kelvin: 4C - 277K and 13C = 286K. I found  e_\text{ideal} = 1 - 277/286= 3.1 %.
    3.  e_\text{ideal} = 1 - 282/320 = 38/320 \approx .12
  33. Assuming the piston is in a constant area cylinder, the change in volume will be \Delta V = A \Delta L where  \Delta L is the distance the piston traveled. Also, we know  P =  F/A \implies F = PA. Since work is  F\Delta L = PA\Delta L = P \Delta V by the change in volume to change in length conversion. So, the work will be P\Delta V = 400\times 10^3 \times 0.04 \times 0.003 \approx 1.48\times 10^4 \text{J}.
  34. The area under the curve is the area of a rectangle with dimensions  P_\text{old} \text{ and }\Delta V and a triangle of height  \Delta P and base  \Delta V . Plugging in, their sum is  .037\times250 + 1/2\times.037\times 150) \approx 1.2\times 10^4 \text{J}.
    1. Shown in figure below.
    2. We use the formula  PV =nRT, where R = 8.35 and n = 1. So, for A:  .6\times 101\times10^3\times.01=8.35T \implies T \approx 73 \text{K}. For B, we have  T = (.25)(101\times10^3)(.01) \approx 30\text{K}. As state C has twice the volume of B at the same pressure, it must have twice the temperature, or 60K.
    3. AB is constant volume, or isochoric. BC is constant pressure, or isobaric.
    4. By convention, when the volume increases, work is done on the gas (BC) during CA, the gas does work. When the volume is constant there is no 'change in distance' and therefore no work is done.
    5. The net work done will be the are of \triangle ABC = 1/2(.6-.25)(101\times10^3)(0.01-0.01) \approx 180 \text{J} .
    1. Shown in figure below.
    2. As shown in the figure, during AB, the pressure doubles. During BC, the volume doubles. Therefore, by the Ideal Gas Law (since the ratio of temperature to the product of pressure and volume remains constant for a constant number of atoms), we have T_C/(P_CV_C) = T_C/(2\times P_A2\times V_A)=T_A/(P_AV_A) \implies T_C = 4T_A . If the gas starts out at STP (300K), it will reach 1200K at its hottest point.
    1. Using the formula  \Delta U + W = Q_\text{net}, we get  Q_\text{net} = 175\text{J}
    2. No work was done from B to C since the volume didn't change. Then, starting from  Q_{AC} = W_{AC} +\Delta U_{AC} I got:  Q_{AC} = Q_{AB}+Q_{BC} =75+(-20) \implies 175 + Q_{BC} =75-20 which allowed me to solve form  Q_{BC} =-120J.
      1.  W_{CD} = -40 . Analogously to above, no work is done from D to A. So,  Q_{CA} =W_{CA}+\Delta U_{CA}\implies -40-\Delta U_{AC} =-20, since the net change in energy over a whole cycle must be 0 (why?).
    3. Work is done only on the constant pressure parts of the cycle, A to B and C to D. Therefore,  W_\text{net} = 75-40=35\text{J}. Over 4 cycles, this is 140 J.
      1. We have  Q_{BD} = Q_{BC}+Q_{CD}=\Delta U_{BD} + W_{BD}. Plugging in, we get  -120+Q_{CD} =-180+(-40) As the change in internal energy from B to D was -180J.
      2. Since the net change in energy over a cycle is 0, 0= \Delta U_{BA} + \Delta U_{AB}= \Delta U_{DA}+100-180, which means that  \Delta U_{DA} = 80\text{J}.
      3. Since work is 0 on constant volume processes,  Q_{DA} =\Delta U_{DA} = 80\text{J}.
    1. We can use the molar version of the Ideal Gas Law to calculate  T = PV/(nR) . So,  T_A = \frac{1.04\times10^{-3}\times 2.5\times 10^5}{.1\times8.315} \approx 313K . Analogously, we find  T_B \approx 500\text{K} , T_C \approx 601 \text{K}, T_D \approx 451 \text{K}.
    2. Shown on diagram.
    3. For isobaric (constant pressure) processes,  W = P\Delta V , therefore  W_{BC} = 4\times10^5(.21\times10^{-3}) =84\text{J}. Likewise,  W_{DA} = 2.5\times10^4(-.46\times10^{-3}) = -115 \text{J} . Since the process from C to D is linear, we can use the average pressure of  \frac{4+2.5}{2} =3.25\text{atm} . Then, using  W_{CD} = P\Delta V = (3.25\times10^4)(.25\times10^3) = 81\text{J}.
    4. SInce the process AB is isochoric, no work is done. So, W_\text{net} = 81+84-115 =50\text{J} .
    5. Over 1 cycle,  \Delta U = 0. Since  \Delta U_{AB}=\Delta U_{CD} =0 , it must hold that  \Delta U_{BC} + \Delta U_{DA} = 0. Therefore,  \Delta U_{BC} = 75\text{J} .
    6.  e = W/Q =50/75 = 2/3 .

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