Ideal Gas Law and Kinetic Theory — Practice Quiz
A Physics cheat sheet for Ideal Gas Law and Kinetic Theory — every key formula with its symbols defined — plus a medium-level practice quiz to test recall.
Formulas & key concepts
Ideal Gas Law: Relates pressure \(P\), volume \(V\), number of moles \(n\), and temperature \(T\).
Where: \(R\) = universal gas constant \((8.31 J/(mol\cdot K))\)
Boyle's Law: For a fixed mass of ideal gas at constant temperature, pressure and volume are inversely proportional.
Charles' Law: For a fixed mass of ideal gas at constant pressure, volume is directly proportional to absolute temperature.
Average Kinetic Energy: Average translational kinetic energy of a gas molecule is proportional to absolute temperature.
Where: \(k\) = Boltzmann constant \((1.38 \times 10^{-23} J/K)\), \(v_{rms}\) = root-mean-square speed
Internal Energy (Monatomic Ideal Gas): Total internal energy \(U\) depends only on temperature.
Fick's Law of Diffusion: Mass \(m\) diffusing in time \(t\) through area \(A\) and length \(L\) due to concentration difference \(\Delta C\).
Where: \(D\) = diffusion constant
Practice quiz
A sealed container holds an ideal gas at a certain pressure and temperature. If the volume of the container is halved while the temperature is kept constant, what happens to the pressure of the gas?
- It doubles.
- It halves.
- It remains the same.
- It quadruples.
Answer: It doubles.
A gas occupies $10 \text{ L}$ at a pressure of $2 \text{ atm}$. If the temperature remains constant, what volume will it occupy if the pressure is increased to $4 \text{ atm}$?
- $2.5 \text{ L}$
- $5 \text{ L}$
- $10 \text{ L}$
- $20 \text{ L}$
Answer: $5 \text{ L}$
A balloon contains $2.0 \text{ L}$ of air at $27 \text{ \textdegree C}$. If the pressure remains constant, what will be its volume if the temperature is increased to $127 \text{ \textdegree C}$?
- $1.5 \text{ L}$
- $2.0 \text{ L}$
- $2.67 \text{ L}$
- $3.0 \text{ L}$
Answer: $2.67 \text{ L}$
According to the kinetic theory of gases, if the absolute temperature of an ideal gas is doubled, what happens to the average translational kinetic energy of its molecules?
- It remains the same.
- It is halved.
- It is doubled.
- It is quadrupled.
Answer: It is doubled.
For a monatomic ideal gas, which of the following factors primarily determines its total internal energy $U$?
- Pressure and volume.
- Number of moles and temperature.
- Volume and temperature.
- Pressure and number of moles.
Answer: Number of moles and temperature.
Which of the following changes would increase the rate of diffusion of a substance according to Fick's Law?
- Decreasing the diffusion constant $D$.
- Increasing the length $L$ of the diffusion path.
- Decreasing the concentration difference $\Delta C$.
- Increasing the cross-sectional area $A$.
Answer: Increasing the cross-sectional area $A$.
An ideal gas is initially at pressure $P_1$, volume $V_1$, and temperature $T_1$. If its volume is compressed to $V_2 = \frac{1}{2} V_1$ and its temperature is increased to $T_2 = 2 T_1$, what is the new pressure $P_2$ in terms of $P_1$?
- $P_2 = P_1$
- $P_2 = 2 P_1$
- $P_2 = 4 P_1$
- $P_2 = \frac{1}{4} P_1$
Answer: $P_2 = 4 P_1$
What is the average translational kinetic energy of a gas molecule at $300 \text{ K}$? Use the Boltzmann constant $k = 1.38 \times 10^{-23} \text{ J/K}$.
- $2.07 \times 10^{-21} \text{ J}$
- $4.14 \times 10^{-21} \text{ J}$
- $6.21 \times 10^{-21} \text{ J}$
- $8.28 \times 10^{-21} \text{ J}$
Answer: $6.21 \times 10^{-21} \text{ J}$
Calculate the internal energy $U$ of $2.0$ moles of a monatomic ideal gas at a temperature of $400 \text{ K}$. Use the universal gas constant $R = 8.31 \text{ J/(mol} \cdot \text{K)}$.
- $4986 \text{ J}$
- $9972 \text{ J}$
- $19944 \text{ J}$
- $2493 \text{ J}$
Answer: $9972 \text{ J}$
A substance diffuses through a membrane with a diffusion constant $D = 2.0 \times 10^{-9} \text{ m}^2/\text{s}$. If the membrane has an area $A = 0.01 \text{ m}^2$ and length $L = 0.001 \text{ m}$, and the concentration difference is $\Delta C = 5.0 \text{ mol/m}^3$, what mass (in moles) diffuses in $100 \text{ s}$?
- $1.0 \times 10^{-5} \text{ mol}$
- $2.0 \times 10^{-5} \text{ mol}$
- $5.0 \times 10^{-6} \text{ mol}$
- $1.0 \times 10^{-6} \text{ mol}$
Answer: $1.0 \times 10^{-5} \text{ mol}$
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