IB Physics Quiz: Understand Gas Laws

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Understand Gas LawsQuestion 1 of 20

The internal energy of a fixed mass of a monatomic ideal gas is $U$ . If the absolute temperature of the gas is halved and the number of moles is doubled, what is the new internal energy of the gas?

U/2

4U

2U

U

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IB Physics Quiz

IB Physics Quiz: Understand Gas Laws

Practice Understand Gas Laws in IB Physics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Understand Gas Laws, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Physics.

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Question 1

The internal energy of a fixed mass of a monatomic ideal gas is $U$ . If the absolute temperature of the gas is halved and the number of moles is doubled, what is the new internal energy of the gas?

$U/2$
$4U$
$2U$
$U$ (correct answer)

Explanation: The internal energy of a monatomic ideal gas is given by

U = \frac{3}{2}nRT

. Let the initial internal energy be

U_1 = \frac{3}{2}n_1RT_1

. The new conditions are

T_2 = T_1/2

and

n_2 = 2n_1

. The new internal energy is

U_2 = \frac{3}{2}n_2RT_2 = \frac{3}{2}(2n_1)R(T_1/2) = \frac{3}{2}n_1RT_1 = U_1

. The internal energy remains unchanged.

Question 2

Under which of the following conditions does a real gas most closely approximate the behavior of an ideal gas?

High pressure and low temperature.
Low pressure and high temperature. (correct answer)
High pressure and high temperature.
Low pressure and low temperature.

Explanation: An ideal gas is modeled as having particles with negligible volume and no intermolecular forces. Real gases approximate this behavior at low pressure, where the molecules are far apart, making their individual volumes negligible compared to the container volume. They also approximate it at high temperature, where the molecules' high kinetic energy is much greater than the potential energy of any intermolecular forces, making those forces insignificant.

Question 3

Two different monatomic ideal gases, X and Y, are mixed in a container and are in thermal equilibrium. The mass of a molecule of gas Y is four times the mass of a molecule of gas X ( $m_Y = 4m_X$ ). What is the ratio of the root-mean-square (RMS) speed of molecules of X to that of molecules of Y ( $v_{rms,X} / v_{rms,Y}$ )?

1/4
1/2
2 (correct answer)
4

Explanation: Since the gases are in thermal equilibrium, they are at the same temperature. The average kinetic energy of the molecules of both gases is the same (

E_k = \frac{3}{2}k_B T

). The kinetic energy is given by

\frac{1}{2}mv_{rms}^2

. Therefore,

\frac{1}{2}m_X v_{rms,X}^2 = \frac{1}{2}m_Y v_{rms,Y}^2

. This simplifies to

v_{rms,X}^2 / v_{rms,Y}^2 = m_Y / m_X

. Given

m_Y = 4m_X

, the ratio is 4. Taking the square root gives

v_{rms,X} / v_{rms,Y} = \sqrt{4} = 2

Question 4

A container of volume 10.0 m³ contains 2.0 moles of a monatomic ideal gas. The total internal energy of the gas is 7.50 × 10³ J. What is the pressure of the gas? (Ideal gas constant R = 8.31 J mol⁻¹ K⁻¹)

250 Pa
500 Pa (correct answer)
750 Pa
1500 Pa

Explanation: For a monatomic ideal gas, the internal energy is U = (3/2)nRT. From the ideal gas law PV = nRT, we can derive that nRT = PV. Substituting this into the internal energy equation gives U = (3/2)PV. Rearranging for pressure: P = 2U/(3V) = (2 × 7.50 × 10³ J)/(3 × 10.0 m³) = 15000/30 = 500 Pa.

Question 5

An ideal gas has a molar mass $M$ . Which expression represents the density $\rho$ of the gas in terms of pressure $P$ , absolute temperature $T$ , and the ideal gas constant $R$ ?

$\frac{RT}{PM}$
$\frac{P}{RTM}$
$\frac{PM}{RT}$ (correct answer)
$\frac{PRT}{M}$

Explanation: Start with the ideal gas law:

PV = nRT

. The number of moles

n

is the total mass

m

divided by the molar mass

M

, so

n = m/M

. Substituting this gives

PV = (m/M)RT

. Density

\rho

is mass per unit volume,

\rho = m/V

. Rearranging the equation to solve for

m/V

gives

m/V = PM/RT

. Therefore,

\rho = PM/RT

Question 6

A cylinder with a movable, frictionless piston contains a fixed amount of an ideal gas at 27 °C. The gas is heated at constant pressure until its volume doubles. What is the final temperature of the gas?

54 °C
273 °C
300 °C
327 °C (correct answer)

Explanation: At constant pressure, Charles's Law applies:

V_1/T_1 = V_2/T_2

. Gas law calculations must use absolute temperature (Kelvin). The initial temperature is

T_1 = 27 + 273 = 300

K. The final volume is

V_2 = 2V_1

. So,

V_1/300 = 2V_1/T_2

. Solving for

T_2

gives

T_2 = 2 \times 300 = 600

K. To convert this back to Celsius, subtract 273:

600 - 273 = 327

°C. Choosing 54 °C is a common error from doubling the Celsius temperature directly.

Question 7

Container X has volume V and holds an ideal gas at pressure P. Container Y has volume 2V and holds the same ideal gas, also at pressure P. Both containers are at the same temperature. A valve connecting them is opened, and the gas reaches equilibrium while the temperature is kept constant. What is the final pressure?

P/2
2P/3 (correct answer)
P
3P/2

Explanation: Initially, container X has nX = PV/(RT) moles and container Y has nY = P(2V)/(RT) moles. The total number of moles is ntotal = PV/(RT) + 2PV/(RT) = 3PV/(RT). After opening the valve, the total volume is V + 2V = 3V. Using PV = nRT for the final state: Pfinal = ntotal RT/(3V) = [3PV/(RT)] × RT/(3V) = P × (3V/3V) = 2P/3.

Question 8

A sealed cubic container with side length $L$ contains an ideal gas at pressure $P$ . What is the average magnitude of the force exerted by the gas on one face of the container?

$P L^2$ (correct answer)
$P L^3$
$6 P L^2$
$P / L^2$

Explanation: Pressure is defined as force per unit area,

P = F/A

. Therefore, the force exerted is

F = P \times A

. The area of one face of a cube with side length

L

A = L^2

. Thus, the average force on one face is

F = P L^2

L^3

is the volume, and

6L^2

is the total surface area of all six faces.

Question 9

A fixed mass of an ideal gas undergoes a process where its pressure is tripled and its volume is reduced to one-third of its initial value. What is the effect on its root-mean-square (RMS) molecular speed?

It remains unchanged. (correct answer)
It is reduced to one-third of its initial value.
It is tripled.
It is multiplied by $\sqrt{3}$ .

Explanation: The RMS speed

v_{rms}

is proportional to the square root of the absolute temperature

T

. We must first determine the change in temperature using the combined gas law:

T_2/T_1 = (P_2/P_1)(V_2/V_1)

. We are given

P_2 = 3P_1

and

V_2 = V_1/3

. Therefore,

T_2/T_1 = (3)(1/3) = 1

. The temperature does not change. Since

v_{rms}

depends only on temperature for a given gas, the RMS speed remains unchanged.

Question 10

A bubble of an ideal gas rises from the bottom of a lake to the surface. Its volume increases by a factor of three. The water temperature is constant throughout the lake. What is the ratio of the pressure at the bottom of the lake to the pressure at the surface?

1/3
1
3 (correct answer)
9

Explanation: Since the temperature is constant and the mass of gas in the bubble is constant, Boyle's Law applies:

P_1V_1 = P_2V_2

. Let state 1 be at the bottom and state 2 be at the surface. We are given

V_2 = 3V_1

. We need to find the ratio

P_1/P_2

. From Boyle's Law,

P_1/P_2 = V_2/V_1

. Substituting the given information,

P_1/P_2 = (3V_1)/V_1 = 3

Question 11

The pressure $P$ of an ideal gas is given by the kinetic theory equation $P = \frac{1}{3}\rho \langle c^2 \rangle$ , where $\rho$ is the gas density and $\langle c^2 \rangle$ is the mean-square speed of the molecules. A fixed mass of ideal gas in a rigid container is heated from 300 K to 900 K. By what factor does the term $\rho \langle c^2 \rangle$ increase?

3 (correct answer)
$\sqrt{3}$
6
9

Explanation: For a fixed mass of gas in a rigid container, both the number of moles

n

and the volume

V

are constant. From the ideal gas law

PV=nRT

, pressure is directly proportional to absolute temperature,

P \propto T

. If the temperature triples (from 300 K to 900 K), the pressure also triples. The kinetic theory equation states

P = \frac{1}{3}\rho \langle c^2 \rangle

, which can be written as

P \propto \rho \langle c^2 \rangle

. Since the pressure triples, the term

\rho \langle c^2 \rangle

must also increase by a factor of 3.

Question 12

The microscopic origin of the pressure exerted by a gas on the walls of its container is the

sum of the gravitational forces of all molecules on the wall.
electrostatic repulsion between molecules and the wall.
average kinetic energy of the molecules within the container.
rate of change of momentum of molecules colliding with the wall. (correct answer)

Explanation: According to the kinetic theory of gases, pressure is a result of the constant collisions of gas molecules with the container walls. Each collision involves a change in the molecule's momentum. By Newton's second law, a change in momentum implies a force exerted on the molecule by the wall, and by Newton's third law, the molecule exerts an equal and opposite force on the wall. The macroscopic pressure is the average force per unit area from these countless collisions, directly related to the rate of change of momentum.

Question 13

A sealed, rigid container holds a fixed mass of an ideal gas. The absolute temperature of the gas is doubled. What is the effect on the pressure of the gas and the average translational kinetic energy of a gas molecule?

The pressure is unchanged, and the average kinetic energy is doubled.
The pressure is doubled, and the average kinetic energy is quadrupled.
The pressure is doubled, and the average kinetic energy is doubled. (correct answer)
The pressure is quadrupled, and the average kinetic energy is doubled.

Explanation: For an ideal gas at constant volume, pressure is directly proportional to absolute temperature (from PV = nRT, since V and n are constant). Therefore, doubling the temperature doubles the pressure. The average translational kinetic energy of gas molecules is given by (3/2)kT, so it is also directly proportional to absolute temperature. Doubling the temperature therefore doubles the average kinetic energy.

Question 14

A tire is inflated with air to a gauge pressure of $2.0 \times 10^5$ Pa when the air temperature is 27 °C. After driving, the air temperature inside the tire rises to 57 °C. Atmospheric pressure is $1.0 \times 10^5$ Pa. What is the new gauge pressure in the tire, assuming its volume remains constant?

$2.3 \times 10^5$ Pa (correct answer)
$2.1 \times 10^5$ Pa
$3.3 \times 10^5$ Pa
$4.0 \times 10^5$ Pa

Explanation: Gas law calculations require absolute pressure and absolute temperature. Initial absolute pressure

P_1 = P_{gauge} + P_{atm} = 2.0 \times 10^5 + 1.0 \times 10^5 = 3.0 \times 10^5

Pa. Initial temperature

T_1 = 27 + 273 = 300

K. Final temperature

T_2 = 57 + 273 = 330

K. For constant volume,

P_1/T_1 = P_2/T_2

. The final absolute pressure is

P_2 = P_1 \times (T_2/T_1) = (3.0 \times 10^5 \text{ Pa}) \times (330 \text{ K} / 300 \text{ K}) = 3.3 \times 10^5

Pa. The question asks for the final gauge pressure:

P_{gauge,2} = P_2 - P_{atm} = 3.3 \times 10^5 - 1.0 \times 10^5 = 2.3 \times 10^5

Pa.

Question 15

A sample of an ideal gas has volume $V$ , pressure $P$ , and absolute temperature $T$ . A second sample of the same ideal gas has twice the volume and twice the pressure, but the same temperature. What is the ratio of the number of molecules in the second sample to the number of molecules in the first sample?

1
2
4 (correct answer)
8

Explanation: The ideal gas law is

PV = N k_B T

, where

N

is the number of molecules. For the first sample,

N_1 = P_1V_1/(k_B T_1)

. For the second sample,

N_2 = P_2V_2/(k_B T_2)

. We are given

P_2 = 2P_1

V_2 = 2V_1

, and

T_2 = T_1

. The ratio is

N_2/N_1 = (P_2V_2/k_B T_2) / (P_1V_1/k_B T_1) = (P_2/P_1)(V_2/V_1)(T_1/T_2)

. Substituting the values:

N_2/N_1 = (2)(2)(1) = 4

Question 16

The pressure of a fixed mass of an ideal gas is doubled, and its volume is halved. What is the ratio of the final absolute temperature to the initial absolute temperature?

1/4
1 (correct answer)
2
4

Explanation: Using the combined gas law,

P_1V_1/T_1 = P_2V_2/T_2

. We are given

P_2 = 2P_1

and

V_2 = V_1/2

. We want to find the ratio

T_2/T_1

. Rearranging the equation gives

T_2/T_1 = (P_2/P_1)(V_2/V_1)

. Substituting the given values:

T_2/T_1 = (2P_1/P_1)( (V_1/2) / V_1) = (2)(1/2) = 1

. The final temperature is the same as the initial temperature.

Question 17

A sealed container of fixed volume contains a mixture of two ideal gases: $2.0 \text{ mol}$ of gas A and $3.0 \text{ mol}$ of gas B. The total pressure is $5.0 \text{ atm}$ at $25°\text{C}$ . If the container is heated until the partial pressure of gas A becomes $4.0 \text{ atm}$ , what is the final temperature?

The final temperature is $477\text{ K}$ or $204°\text{C}$
The final temperature is $372\text{ K}$ or $99°\text{C}$
The final temperature is $596\text{ K}$ or $323°\text{C}$ (correct answer)
The final temperature is $298\text{ K}$ or $25°\text{C}$

Explanation: When you encounter gas mixture problems involving temperature changes, you need to apply Dalton's Law of Partial Pressures combined with Gay-Lussac's Law. The key insight is that partial pressure is proportional to both the number of moles and temperature. First, find gas A's initial partial pressure. Since partial pressure depends on mole fraction:

P_A = \frac{n_A}{n_{total}} \times P_{total} = \frac{2.0}{5.0} \times 5.0 = 2.0 \text{ atm}

For a fixed volume and amount of gas, partial pressure is directly proportional to absolute temperature:

\frac{P_1}{T_1} = \frac{P_2}{T_2}

Converting the initial temperature:

T_1 = 25°C + 273 = 298 K

Solving for final temperature:

T_2 = T_1 \times \frac{P_2}{P_1} = 298 K \times \frac{4.0}{2.0} = 596 K

Converting back:

596 - 273 = 323°C

Answer C (

596 K

323°C

) is correct. Answer A (

477 K

) incorrectly assumes some other pressure ratio. Answer B (

372 K

) might result from using the wrong initial partial pressure or incorrect temperature conversion. Answer D (

298 K

) represents no temperature change, ignoring that pressure doubled. Remember: In gas mixture problems, always calculate partial pressures using mole fractions first, then apply gas laws to individual components. The partial pressure of one gas changing tells you about the overall temperature change affecting the entire mixture.

Question 18

A gas undergoes a cyclic process consisting of three steps: (1) isobaric expansion from volume $V$ to $2V$ at pressure $P$ , (2) isochoric cooling back to the original temperature, and (3) isothermal compression back to the original state. What is the pressure after step 2?

The pressure after step 2 is $\frac{P}{4}$
The pressure after step 2 is $P$
The pressure after step 2 is $2P$
The pressure after step 2 is $\frac{P}{2}$ (correct answer)

Explanation: When analyzing cyclic thermodynamic processes, you need to track how pressure, volume, and temperature change through each step using the ideal gas law:

PV = nRT

. Let's trace through each step systematically. Initially, we have state (

P

V

T_0

). In step 1, isobaric expansion doubles the volume at constant pressure

P

. Since

PV = nRT

, when volume doubles at constant pressure, temperature must also double to

2T_0

. So after step 1: (

P

2V

2T_0

). Step 2 is isochoric cooling "back to the original temperature"

T_0

at constant volume

2V

. Using the ideal gas law again: if temperature halves while volume stays constant, pressure must also halve. Therefore:

P_2 = \frac{P}{2}

at state (

\frac{P}{2}

2V

T_0

). Now let's examine the wrong answers. Answer A (

\frac{P}{4}

) would occur if you incorrectly thought both volume and temperature changes affected pressure multiplicatively. Answer B (

P

) assumes pressure somehow returns to its original value, ignoring that we're at a different volume-temperature combination. Answer C (

2P

) incorrectly suggests pressure increases during cooling at constant volume. The key insight is that answer D (

\frac{P}{2}

) correctly applies the ideal gas law: when temperature decreases by half at constant volume, pressure must decrease proportionally. Study tip: For thermodynamic cycles, always track all three variables (P, V, T) through each step using

PV = nRT

. Draw a table or diagram to avoid losing track of the state changes.

Question 19

An ideal gas in a cylinder with a movable piston undergoes the process shown in the P-V diagram. The gas starts at point A ( $2.0 \text{ atm}$ , $1.0 \text{ L}$ ), undergoes isobaric expansion to point B ( $2.0 \text{ atm}$ , $3.0 \text{ L}$ ), then isochoric cooling to point C ( $P_C$ , $3.0 \text{ L}$ ). If the temperature at point A is $300 \text{ K}$ , what is the pressure at point C?

The pressure at point C is $1.33 \text{ atm}$
The pressure at point C is $0.67 \text{ atm}$ (correct answer)
The pressure at point C is $6.0 \text{ atm}$
The pressure at point C is $1.0 \text{ atm}$

Explanation: First find temperature at B using Charles's law (isobaric process):

\frac{V_A}{T_A} = \frac{V_B}{T_B}

, so

T_B = T_A \times \frac{V_B}{V_A} = 300 \times \frac{3.0}{1.0} = 900\text{ K}

. For isochoric process B→C, the gas returns to original temperature

T_C = T_A = 300\text{ K}

. Using Gay-Lussac's law:

\frac{P_B}{T_B} = \frac{P_C}{T_C}

, so

P_C = P_B \times \frac{T_C}{T_B} = 2.0 \times \frac{300}{900} = 0.67\text{ atm}

Question 20

A sealed container holds an ideal gas at temperature $T_1 = 300 \text{ K}$ and pressure $P_1 = 2.0 \times 10^5 \text{ Pa}$ . The gas is heated until its pressure doubles, then expanded isothermally until its pressure returns to the original value. What is the final volume of the gas compared to its initial volume?

The final volume is twice the initial volume (correct answer)
The final volume is four times the initial volume
The final volume is equal to the initial volume
The final volume is half the initial volume

Explanation: Using the ideal gas law for each step: Initially

PV = nRT_1

. After heating at constant volume:

P_2 = 2P_1

T_2 = 2T_1 = 600\text{ K}

. During isothermal expansion:

P_2V_2 = P_1V_f

where

V_f

is final volume. Since

P_2 = 2P_1

2P_1V_2 = P_1V_f

, so

V_f = 2V_2 = 2V_1

. The final volume is twice the initial volume.