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Gas Pressure and Force

Pressure is the force exerted over a given area and the behavior of most gases can be described using the ideal gas law.

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Gas Pressure and Force

Students will learn the basics of pressure and force, how pressure varies with altitude in our atmosphere, and the ideal gas law, which connects the concepts of volume, pressure, temperature, and the number of particles that compose a gas.

Key Equations

\begin{align*}P = F / A && \text{pressure equals the force divided by the area of application} \\ P = P_0 e^\frac{-h} {a} && \text{exponential model for atmospheric pressure}\end{align*}P=F/AP=P0ehapressure equals the force divided by the area of applicationexponential model for atmospheric pressure

\begin{align*} PV = NkT && \text{Ideal Gas Law}\end{align*}PV=NkTIdeal Gas Law

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 \begin{align*}N\end{align*}N molecules of such a gas at temperature \begin{align*}T\end{align*}T and volume \begin{align*}V\end{align*}V, the pressure can be calculated from this formula. Note that \begin{align*}k=1.38\times10^{-23}\;\mathrm{J/K}\end{align*}k=1.38×1023J/K

  • 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 (\begin{align*}1 \;\mathrm{atm} = 101,000\end{align*}1atm=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.

Example 1

How many molecules of gas does it take to equalize the pressure inside a 1 liter box, that originally starts with no gas inside, with the atmospheric pressure at room temperature (300 K)?


To solve this problem, we just plug in the given values into the ideal gas law.

\begin{align*} PV&=NkT\\ N&=\frac{PV}{kT}\\ N&=\frac{101000\;\text{Pa} * 1\;\text{L}}{1.38*10^{-23}\;\text{J/K} * 300\;\text{K}}\\ N&=2.44*10^{25}\;\text{molecules}\\ \end{align*}PVNNN=NkT=PVkT=101000Pa1L1.381023J/K300K=2.441025molecules

Watch this Explanation

Time for Practice

  1. 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.
  2. Use the formula \begin{align*}P = F / A\end{align*}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).
  3. 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?
  4. 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?
  5. Why is it a good idea for Noreen to open her bag of chips before she drives to the top of a high mountain?
  6. 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.)
  7. Calculate the amount of force exerted on a \begin{align*}2 \;\mathrm{cm} \times 2 \;\mathrm{cm}\end{align*}2cm×2cm patch of your skin due to atmospheric pressure \begin{align*}(P_0 = 101,000 \;\mathrm{Pa})\end{align*}(P0=101,000Pa). Why doesn’t your skin burst under this force?
  8. Assuming that the pressure of the atmosphere decreases exponentially as you rise in elevation according to the formula \begin{align*}P = P_0 e^\frac{-h} {a}\end{align*}P=P0eha, where \begin{align*}P_0\end{align*}P0 is the atmospheric pressure at sea level \begin{align*}(101,000 \;\mathrm{Pa})\end{align*}(101,000Pa), \begin{align*}h\end{align*}h is the altitude in km and a is the scale height of the atmosphere \begin{align*}(a \approx 8.4 \;\mathrm{km})\end{align*}(a8.4km).
    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 \begin{align*}2 \;\mathrm{km}\end{align*}2km 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?
  9. An instructor has an ideal monatomic helium gas sample in a closed container with a volume of \begin{align*}0.01\;\mathrm{m}^3\end{align*}0.01m3, a temperature of \begin{align*}412 \;\mathrm{K}\end{align*}412K, and a pressure of \begin{align*}474 \;\mathrm{kPa}\end{align*}474kPa.
    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 \begin{align*}647 \;\mathrm{K}\end{align*}647K. 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.

Answers to Selected Problems

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  7. \begin{align*}40 \;\mathrm{N}\end{align*}40N
  8. a. \begin{align*}21,000 \;\mathrm{Pa}\end{align*}21,000Pa b. Decreases to \begin{align*}61,000 \;\mathrm{Pa}\end{align*}61,000Pa c. \begin{align*}5.8 \;\mathrm{km}\end{align*}5.8km
  9. a.\begin{align*} 8.34 \times 10^{23}\end{align*}8.34×1023 b. \begin{align*}6.64 \times 10^{-27}\;\mathrm{kg}\end{align*}6.64×1027kg c. \begin{align*}1600 \;\mathrm{m/s}\end{align*}1600m/s d. \begin{align*}744 \;\mathrm{kPa}\end{align*}744kPa e. \begin{align*}4.2 \times 10^{20}\end{align*}4.2×1020 or \begin{align*}0.0007 \;\mathrm{moles}\end{align*}0.0007moles g. \begin{align*}0.00785 \;\mathrm{m}^3\end{align*}0.00785m3

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