<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />

# 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.

Estimated10 minsto complete
%
Progress
Practice Gas Pressure and Force

MEMORY METER
This indicates how strong in your memory this concept is
Progress
Estimated10 minsto complete
%
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

P=F/AP=P0ehapressure equals the force divided by the area of applicationexponential model for atmospheric pressure\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*}

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

Guidance
• 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 (1atm=101,000\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.

PVNNN=NkT=PVkT=101000Pa1L1.381023J/K300K=2.441025molecules\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*}

### Time for Practice

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

#### Answers to Selected Problems

1. 40N\begin{align*}40 \;\mathrm{N}\end{align*}
2. a. 21,000Pa\begin{align*}21,000 \;\mathrm{Pa}\end{align*} b. Decreases to 61,000Pa\begin{align*}61,000 \;\mathrm{Pa}\end{align*} c. 5.8km\begin{align*}5.8 \;\mathrm{km}\end{align*}
3. a.8.34×1023\begin{align*} 8.34 \times 10^{23}\end{align*} b. 6.64×1027kg\begin{align*}6.64 \times 10^{-27}\;\mathrm{kg}\end{align*} c. 1600m/s\begin{align*}1600 \;\mathrm{m/s}\end{align*} d. 744kPa\begin{align*}744 \;\mathrm{kPa}\end{align*} e. 4.2×1020\begin{align*}4.2 \times 10^{20}\end{align*} or 0.0007moles\begin{align*}0.0007 \;\mathrm{moles}\end{align*} g. 0.00785m3\begin{align*}0.00785 \;\mathrm{m}^3\end{align*}

### Notes/Highlights Having trouble? Report an issue.

Color Highlighted Text Notes
Please to create your own Highlights / Notes
Show More

### Explore More

Sign in to explore more, including practice questions and solutions for Gas Pressure and Force.
Please wait...
Please wait...