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*}

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

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*} 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*}

- 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.

#### 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)?

##### Solution

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*}

### Watch this Explanation

### Time for Practice

- 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?

#### Answers to Selected Problems

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