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*} 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.
An ideal gas is a gas where the atoms are treated as pointparticles and assumed to never collide or interact with each other. If you have \begin{align*}N\end{align*} molecules of such a gas at temperature \begin{align*}T\end{align*} and volume \begin{align*}V\end{align*}, the pressure can be calculated from this formula. Note that \begin{align*}k=1.38\times10^{23}\;\mathrm{J/K}\end{align*}
\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*}
\begin{align*} PV = NkT && \text{Ideal Gas Law}\end{align*}
Example
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*}
Interactive Simulation
Review
 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.
 Use the formula \begin{align*}P = F / A\end{align*} to argue why it is easier to pop a balloon with a needle than with a finger (pretend you don’t have long fingernails).
 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?
 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?
 Why is it a good idea for Noreen to open her bag of chips before she drives to the top of a high mountain?
 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.)
 Calculate the amount of force exerted on a \begin{align*}2 \;\mathrm{cm} \times 2 \;\mathrm{cm}\end{align*} patch of your skin due to atmospheric pressure \begin{align*}(P_0 = 101,000 \;\mathrm{Pa})\end{align*}. Why doesn’t your skin burst under this force?
 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*}, where \begin{align*}P_0\end{align*} is the atmospheric pressure at sea level \begin{align*}(101,000 \;\mathrm{Pa})\end{align*}, \begin{align*}h\end{align*} is the altitude in km and a is the scale height of the atmosphere \begin{align*}(a \approx 8.4 \;\mathrm{km})\end{align*}.
 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*} above sea level.
 If you rise to half the scale height of Earth’s atmosphere, by how much does the pressure decrease?
 If the pressure is half as much as on sea level, what is your elevation?
 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*}, a temperature of \begin{align*}412 \;\mathrm{K}\end{align*}, and a pressure of \begin{align*}474 \;\mathrm{kPa}\end{align*}.
 Approximately how many gas atoms are there in the container?
 Calculate the mass of the individual gas atoms.
 Calculate the speed of a typical gas atom in the container.
 The container is heated to \begin{align*}647 \;\mathrm{K}\end{align*}. What is the new gas pressure?
 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?
 Is this number half the number from part (a)? Why or why not?
 The closed container is now compressed isothermally so that the pressure rises to its original pressure. What is the new volume of the container?
 Sketch this process on a PV diagram.
 Sketch cubes with volumes corresponding to the old and new volumes.
Review (Answers)

There will be more collisions because there is an increase in the number of particles in the box; therefore, the pressure will increase.

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.

The bottle crumples since it can no longer push back against atmospheric pressure.

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. 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.

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 at the top of a high mountain. This would make it more difficult to open without spilling.

Answers will vary.

40.4 N . Your skin does not burst because there’s an equal amount of pressure outward (blood pressure).

a. −2.1 × 10^{4} Pa b. −3.9 × 10^{4} Pa c. 5822.4 m

a. 8.34×10^{23} atoms b. 6.64 x 10−27 kg c. 1608 m/s d. 744 kPa e. 4.2×10^{20} atoms f. Yes, the number of atoms didn’t change in the heating. g. 0.00785 m^{3}