AP Physics 2 Quiz: The Ideal Gas Law

20 questions · exam conditions

0:00

The Ideal Gas LawQuestion 1 of 20

A fixed amount of ideal gas is taken from state 1 to state 2 in a cylinder. The temperature is held constant at $300\,\text{K}$ while the volume increases from $1.5\,\text{L}$ to $3.0\,\text{L}$ . If $P_1=300\,\text{kPa}$ , then $P_2$ is

600\,\text{kPa}

because pressure increases when volume increases.

450\,\text{kPa}

because

300\times(3.0/1.5)=450

300\,\text{kPa}

because temperature is constant.

150\,\text{kPa}

because pressure is inversely proportional to volume.

← Back to quizzes

AP Physics 2 Quiz

AP Physics 2 Quiz: The Ideal Gas Law

Practice The Ideal Gas Law in AP Physics 2 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 The Ideal Gas Law, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Physics 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

$600\,\text{kPa}$ because pressure increases when volume increases.
$450\,\text{kPa}$ because $300\times(3.0/1.5)=450$ .
$300\,\text{kPa}$ because temperature is constant.
$150\,\text{kPa}$ because pressure is inversely proportional to volume. (correct answer)

Explanation: This question tests the ideal gas law. At constant temperature and amount of gas, the ideal gas law (PV = nRT) reduces to Boyle's Law: P₁V₁ = P₂V₂, showing that pressure and volume are inversely proportional. When volume doubles from 1.5 L to 3.0 L, pressure must be halved to maintain the constant product PV. Therefore, pressure decreases from 300 kPa to 150 kPa. Choice A incorrectly states that pressure increases when volume increases, revealing a fundamental misconception about the inverse relationship between P and V at constant temperature. When temperature is constant, remember that pressure and volume change in opposite directions by inverse factors.

Question 2

An ideal gas in a cylinder with a movable piston is compressed slowly from $4.0\,\text{L}$ to $2.0\,\text{L}$ while the temperature and amount of gas are held constant. Initially the pressure is $100\,\text{kPa}$ . Compared to the initial pressure, the final pressure is

$50\,\text{kPa}$ because pressure decreases when volume decreases.
$100\,\text{kPa}$ because temperature is constant.
$150\,\text{kPa}$ because pressure increases by $50\,\text{kPa}$ .
$200\,\text{kPa}$ because pressure is inversely proportional to volume. (correct answer)

Explanation: This question tests the ideal gas law. The ideal gas law (PV = nRT) shows that when temperature and amount of gas are constant, pressure and volume are inversely proportional (P₁V₁ = P₂V₂). As volume decreases from 4.0 L to 2.0 L (halved), pressure must double to maintain the same product PV. Therefore, pressure increases from 100 kPa to 200 kPa. Choice A incorrectly states that pressure decreases when volume decreases, revealing a fundamental misconception about the inverse relationship between P and V. To solve ideal gas problems correctly, identify which variables are constant and apply the appropriate proportionality relationship.

Question 3

A fixed amount of ideal gas is kept at constant temperature in a piston-cylinder device. The pressure is reduced from $300\,\text{kPa}$ to $150\,\text{kPa}$ while the amount of gas is constant. Compared to the initial volume, the final volume is

half as large because volume is proportional to pressure.
the same because temperature is constant.
twice as large because volume is inversely proportional to pressure. (correct answer)
four times as large because pressure was reduced by a factor of 2.

Explanation: This question tests the ideal gas law. At constant temperature with a fixed amount of gas, pressure and volume are inversely proportional according to Boyle's Law (P₁V₁ = P₂V₂). When pressure is halved from 300 kPa to 150 kPa, volume must double to maintain the constant product PV. Therefore, the final volume is twice the initial volume. Choice A incorrectly states that volume is proportional to pressure, confusing direct and inverse relationships. Remember that at constant temperature, P and V vary inversely—when one doubles, the other halves.

Question 4

A rigid, sealed $2.0\,\text{L}$ container holds an ideal gas at $1.0\,\text{atm}$ and $300\,\text{K}$ . The amount of gas and volume are held constant while the gas is heated to $450\,\text{K}$ . Which statement correctly describes the final pressure?

It decreases to about $0.67\,\text{atm}$ because temperature increased.
It increases to about $1.5\,\text{atm}$ because pressure is proportional to Kelvin temperature. (correct answer)
It stays at $1.0\,\text{atm}$ because volume is constant.
It increases to about $2.5\,\text{atm}$ because $450-300=150$ .

Explanation: This question tests the ideal gas law. The ideal gas law states that PV = nRT, where pressure (P), volume (V), amount of gas (n), and temperature (T) are related through the gas constant R. When volume and amount of gas are constant, pressure is directly proportional to absolute temperature (P₁/T₁ = P₂/T₂). Since temperature increases from 300 K to 450 K (a factor of 1.5), pressure must also increase by a factor of 1.5, from 1.0 atm to 1.5 atm. Choice D incorrectly adds the temperature difference (150 K) to the pressure, showing a misconception about how to apply proportional relationships. When solving ideal gas problems, always use absolute temperature in Kelvin and identify which variables remain constant to determine the appropriate relationship.

Question 5

An ideal gas in a sealed, rigid container (constant volume and constant amount) is warmed from $27^\circ\text{C}$ to $127^\circ\text{C}$ . Which statement correctly describes the final pressure compared with the initial pressure?

$P_2=P_1$ because rigid containers keep pressure constant
$P_2=\tfrac{300}{400}P_1$ because pressure decreases as temperature increases
$P_2=\tfrac{400}{300}P_1$ because pressure is proportional to kelvin temperature (correct answer)
$P_2=\tfrac{127}{27}P_1$ because pressure is proportional to Celsius temperature

Explanation: This question tests the ideal gas law. For constant volume and amount, pressure is proportional to absolute temperature: P₁/T₁ = P₂/T₂. First convert temperatures to Kelvin: T₁ = 27°C + 273 = 300 K and T₂ = 127°C + 273 = 400 K. Then P₂ = P₁ × (T₂/T₁) = P₁ × (400 K / 300 K) = (4/3)P₁. Choice A incorrectly uses Celsius temperatures directly (127/27), demonstrating the common misconception that temperature ratios work with any scale. Always convert to Kelvin before using temperature ratios in gas law calculations.

Question 6

An ideal gas in a cylinder is kept at constant pressure by a movable piston. Initially $V_1=4.0\,\text{L}$ at $T_1=250\,\text{K}$ . It is warmed to $T_2=300\,\text{K}$ while pressure and amount of gas remain constant. Which statement correctly describes $V_2$ ?

It is $4.8\,\text{L}$ . (correct answer)
It is $3.3\,\text{L}$ .
It is $4.0\,\text{L}$ .
It is $54\,\text{L}$ because $\Delta T=50^\circ\text{C}$ .

Explanation: This question tests the ideal gas law. The ideal gas law PV = nRT shows how pressure, volume, temperature, and moles are interconnected. When pressure and moles are constant, volume is directly proportional to temperature: V₁/T₁ = V₂/T₂. With V₁ = 4.0 L, T₁ = 250 K, and T₂ = 300 K, we calculate V₂ = V₁T₂/T₁ = (4.0 L)(300 K)/(250 K) = 4.8 L. Choice D incorrectly uses the temperature difference in Celsius (50°C) rather than the ratio of absolute temperatures. Always convert to Kelvin and use ratios when applying Charles's Law: V₁/T₁ = V₂/T₂.

Question 7

A flexible balloon contains ideal gas at constant temperature and external pressure. The balloon initially has $n_1=0.20\,\text{mol}$ and volume $V_1=1.0\,\text{L}$ . Gas is added to reach $n_2=0.30\,\text{mol}$ while temperature and pressure remain constant. Compared to $V_1$ , what is $V_2$ ?

It is $1.3\,\text{L}$ because volume depends on Celsius temperature.
It is $0.67\,\text{L}$ .
It is $1.5\,\text{L}$ . (correct answer)
It is $1.0\,\text{L}$ .

Explanation: This question tests the ideal gas law. According to PV = nRT, pressure, volume, temperature, and moles of gas are related in specific ways. When pressure and temperature are constant, volume is directly proportional to the number of moles: V₁/n₁ = V₂/n₂. With V₁ = 1.0 L, n₁ = 0.20 mol, and n₂ = 0.30 mol, we find V₂ = V₁n₂/n₁ = (1.0 L)(0.30 mol)/(0.20 mol) = 1.5 L. Choice D incorrectly suggests using Celsius temperature, which is irrelevant when temperature is constant. When P and T are constant, use Avogadro's Law: V₁/n₁ = V₂/n₂.

Question 8

A rigid, sealed steel tank contains $1.0\ \text{mol}$ of an ideal gas at $300\ \text{K}$ and pressure $P_1$ . The tank is heated to $450\ \text{K}$ while volume and amount of gas remain constant. Which statement correctly describes the new pressure $P_2$ compared to $P_1$ ?

$P_2=\tfrac{2}{3}P_1$
$P_2=1.5P_1$ (correct answer)
$P_2=P_1+150\ \text{kPa}$
$P_2=\tfrac{450}{300+273}P_1$

Explanation: This problem tests understanding of the ideal gas law. The ideal gas law states that PV = nRT, where pressure (P), volume (V), and temperature (T) are related through the number of moles (n) and gas constant (R). When volume and amount of gas are constant, pressure is directly proportional to absolute temperature: P₁/T₁ = P₂/T₂. Since temperature increases from 300 K to 450 K, we get P₂ = P₁ × (450/300) = 1.5P₁. Choice D incorrectly uses Celsius temperature (450/(300+273)), showing the common misconception of not recognizing that temperatures are already in Kelvin. Always verify that temperatures are in absolute units (Kelvin) and identify which variables remain constant to determine the appropriate gas law relationship.

Question 9

A sample of ideal gas is kept at constant pressure in a piston. Its temperature increases from $300\,\text{K}$ to $360\,\text{K}$ , and no gas is added or removed. Which statement correctly describes how the volume changes?

The volume stays the same because pressure is constant.
The volume decreases by a factor of $360/300$ because higher temperature compresses the gas.
The volume increases by $60\%$ because $360-300=60$ .
The volume increases by a factor of $360/300$ because volume is proportional to absolute temperature at constant pressure. (correct answer)

Explanation: This question tests understanding of the ideal gas law. The ideal gas law shows that at constant pressure and amount of gas, volume is directly proportional to absolute temperature (Charles's Law): V₁/T₁ = V₂/T₂. The volume increases by the same factor as the temperature: V₂/V₁ = T₂/T₁ = 360 K / 300 K = 1.2, meaning volume increases by a factor of 360/300. Choice D incorrectly calculates the change as 60% by subtracting temperatures (360 - 300 = 60), showing a misconception about using temperature ratios rather than differences. Always use ratios of absolute temperatures, not temperature differences, when applying gas laws.

Question 10

A fixed amount of ideal gas is in a cylinder with a frictionless piston. The gas is cooled from $T_1=500\ \text{K}$ to $T_2=250\ \text{K}$ while the external pressure is adjusted so the gas pressure remains constant. No gas enters or leaves. Which statement correctly describes $V_2$ compared with $V_1$ ?

$V_2=V_1$ because constant pressure implies constant volume
$V_2=\tfrac{250-273}{500-273}V_1$ because Celsius temperatures must be used
$V_2=2V_1$ because volume increases when temperature decreases at constant pressure
$V_2=\tfrac{1}{2}V_1$ because volume is proportional to kelvin temperature at constant pressure (correct answer)

Explanation: This question tests the ideal gas law. For an ideal gas with constant pressure and constant amount, PV = nRT simplifies to show that volume is directly proportional to absolute temperature: V₁/T₁ = V₂/T₂. Since temperature decreases from 500 K to 250 K (halved), the volume must also be halved: V₂ = V₁ × (250 K / 500 K) = ½V₁. Choice A incorrectly states that volume increases when temperature decreases, revealing a fundamental misconception about the direct relationship between V and T. When pressure and amount are constant, use V₁/T₁ = V₂/T₂ with Kelvin temperatures.

Question 11

A fixed amount of ideal gas is sealed in a cylinder with a movable piston. Initially $P_1=200\ \text{kPa}$ , $V_1=3.0\ \text{L}$ , and $T_1=300\ \text{K}$ . The gas is slowly heated while the piston moves so the pressure is held constant at $200\ \text{kPa}$ and no gas enters or leaves. When the temperature reaches $T_2=450\ \text{K}$ , which statement correctly describes the final volume $V_2$ ?

$V_2=2.0\ \text{L}$ because volume decreases as temperature increases
$V_2=4.5\ \text{L}$ because volume is proportional to temperature in kelvins (correct answer)
$V_2=6.75\ \text{L}$ because volume is proportional to Celsius temperature
$V_2=3.0\ \text{L}$ because pressure is constant so volume stays constant

Explanation: This question tests the ideal gas law. The ideal gas law states that PV = nRT, where pressure (P), volume (V), and temperature (T) are related for a fixed amount of gas (n moles). Since the pressure is held constant and the amount of gas is fixed, we can use Charles's Law: V₁/T₁ = V₂/T₂. Substituting the given values: 3.0 L / 300 K = V₂ / 450 K, which gives V₂ = 3.0 L × (450 K / 300 K) = 4.5 L. Choice C incorrectly uses Celsius temperatures without converting to Kelvin, which would give 6.75 L—this is a common misconception that temperature ratios work with any scale. When pressure and amount are constant, always use absolute temperature (Kelvin) and apply V₁/T₁ = V₂/T₂.

Question 12

A container of fixed volume holds an ideal gas at $P_1=90\ \text{kPa}$ and $T_1=300\ \text{K}$ . The temperature is kept constant while some gas escapes so the number of moles decreases to $\tfrac{2}{3}$ of its original value. Which statement correctly describes the final pressure?

$P_2=30\ \text{kPa}$ , because pressure drops with the square of the moles
$P_2=135\ \text{kPa}$ , because losing gas increases pressure in a rigid container
$P_2=90\ \text{kPa}$ , because pressure depends only on temperature
$P_2=60\ \text{kPa}$ , because pressure is proportional to moles at fixed $V$ and $T$ (correct answer)

Explanation: This problem tests understanding of the ideal gas law. At constant volume and temperature, pressure is directly proportional to the number of moles: P ∝ n. If the number of moles decreases to 2/3 of its original value, the pressure will also decrease to 2/3 of its original value. Therefore, P₂ = (2/3) × 90 kPa = 60 kPa. Choice C incorrectly suggests that losing gas increases pressure, which violates the direct relationship between pressure and amount of gas. When volume and temperature are constant, pressure changes proportionally with the number of moles—removing gas always decreases pressure.

Question 13

An ideal gas in a cylinder has $P_1=1.0\ \text{atm}$ , $V_1=4.0\ \text{L}$ , and $T_1=300\ \text{K}$ . The amount of gas is constant. The gas is heated to $T_2=600\ \text{K}$ while the pressure is held constant. Which statement correctly describes the final volume?

$V_2=4.0\ \text{L}$ , because constant pressure means constant volume
$V_2=8.0\ \text{L}$ , because volume is proportional to absolute temperature (correct answer)
$V_2=2.0\ \text{L}$ , because volume decreases when temperature increases
$V_2=12\ \text{L}$ , because volume triples when temperature doubles

Explanation: This problem tests understanding of the ideal gas law. At constant pressure and fixed amount of gas, Charles's Law applies: V₁/T₁ = V₂/T₂. The temperature doubles from 300 K to 600 K, so the volume must also double to maintain constant pressure. Therefore, V₂ = 4.0 L × (600 K/300 K) = 8.0 L. Choice D incorrectly suggests volume triples when temperature doubles, misunderstanding the direct proportionality. When pressure is constant, always remember that volume and absolute temperature are directly proportional for an ideal gas.

Question 14

A balloon contains an ideal gas at constant temperature. The balloon's volume increases from $2.0\,\text{L}$ to $3.0\,\text{L}$ while the amount of gas remains constant. Which statement correctly describes the change in pressure?

The pressure stays the same because temperature is constant.
The pressure increases by a factor of $3/2$ because pressure is proportional to volume.
The pressure decreases to $2/3$ of its initial value because pressure is inversely proportional to volume. (correct answer)
The pressure decreases by $1.0\,\text{atm}$ because the volume increased by $1.0\,\text{L}$ .

Explanation: This question tests understanding of the ideal gas law. At constant temperature and amount of gas, pressure and volume are inversely proportional (Boyle's Law): P₁V₁ = P₂V₂. When volume increases from 2.0 L to 3.0 L (a factor of 3/2), pressure decreases by the inverse factor: P₂ = P₁ × (V₁/V₂) = P₁ × (2.0 L / 3.0 L) = (2/3)P₁. Choice A incorrectly states pressure is proportional to volume, showing a misconception about the inverse relationship in Boyle's Law. Remember that at constant temperature, pressure and volume are inversely proportional—when one increases, the other decreases.

Question 15

An ideal gas in a piston is compressed so that its pressure doubles while its temperature is held constant and no gas leaks. Which statement correctly describes the final volume compared to the initial volume?

The final volume is unchanged because temperature is constant.
The final volume is one-fourth the initial volume because doubling pressure halves temperature.
The final volume is twice the initial volume because pressure and volume increase together.
The final volume is half the initial volume because volume is inversely proportional to pressure at constant temperature. (correct answer)

Explanation: This question tests understanding of the ideal gas law. At constant temperature with no gas leaking, Boyle's Law applies: P₁V₁ = P₂V₂, showing pressure and volume are inversely proportional. If pressure doubles (P₂ = 2P₁), then volume must be halved: V₂ = V₁ × (P₁/P₂) = V₁ × (1/2) = V₁/2. Choice A incorrectly assumes pressure and volume increase together, demonstrating a fundamental misconception about their inverse relationship. When temperature is constant, remember that pressure and volume move in opposite directions—doubling one halves the other.

Question 16

An ideal gas occupies volume $V$ at pressure $P$ and temperature $T$ . The temperature is doubled to $2T$ while pressure is held constant and no gas is added or removed. Which statement correctly describes the new volume?

$V_2=V_1$
$V_2=\tfrac{2T-273}{T}V_1$
$V_2=2V_1$ (correct answer)
$V_2=\tfrac{1}{2}V_1$

Explanation: This problem tests understanding of the ideal gas law. The ideal gas law PV = nRT shows that pressure, volume, temperature, and amount of gas are all related. When pressure and amount of gas remain constant, volume is directly proportional to absolute temperature: V₁/T₁ = V₂/T₂. If temperature doubles from T to 2T, volume also doubles: V₂ = V₁ × (2T/T) = 2V₁. Choice D incorrectly attempts to subtract 273 from the temperature ratio, showing confusion between Kelvin conversion and temperature ratios. When using the ideal gas law, work with temperature ratios directly if temperatures are already in Kelvin.

Question 17

A cylinder of ideal gas has a movable piston. The gas expands from volume $V_1$ to $2V_1$ while temperature and amount of gas remain constant. Which statement correctly describes the final pressure $P_2$ compared to $P_1$ ?

$P_2=2P_1$
$P_2=P_1$
$P_2=\tfrac{1}{2}P_1$ (correct answer)
$P_2=\tfrac{T_2}{T_1}P_1$ even though $T$ is constant

Explanation: This problem tests understanding of the ideal gas law. The ideal gas law relates pressure, volume, temperature, and amount of gas through PV = nRT. When temperature and amount of gas remain constant, pressure and volume are inversely proportional: P₁V₁ = P₂V₂. Since volume doubles from V₁ to 2V₁, the pressure must halve: P₂ = P₁ × (V₁/2V₁) = ½P₁. Choice A shows the misconception of thinking pressure doubles when volume doubles, failing to recognize the inverse relationship. To solve gas law problems correctly, first identify which variables are constant, then apply the appropriate form of the ideal gas law.

Question 18

A balloon contains an ideal gas at $P=1.0\ \text{atm}$ and $T=300\ \text{K}$ . The external pressure stays at $1.0\ \text{atm}$ , so the gas pressure is constant. If the number of moles inside doubles while temperature remains constant, which statement correctly describes the balloon's volume?

The volume stays the same, because volume depends only on temperature
The volume quadruples, because doubling moles doubles both pressure and volume
The volume halves, because adding gas increases pressure and shrinks the balloon
The volume doubles, because $V$ is proportional to $n$ at constant $P$ and $T$ (correct answer)

Explanation: This problem tests understanding of the ideal gas law. At constant pressure and temperature, the ideal gas law shows that volume is directly proportional to the number of moles: V ∝ n. Since the balloon maintains equilibrium with external pressure (1.0 atm), its internal pressure stays constant. When the number of moles doubles while P and T remain constant, the volume must also double to satisfy PV = nRT. Choice C incorrectly assumes adding gas shrinks the balloon, misunderstanding that constant pressure allows expansion. For problems involving flexible containers like balloons, recognize that pressure equilibrates with surroundings, making volume proportional to the amount of gas.

Question 19

An ideal gas sample has $P_1=150\,\text{kPa}$ , $V_1=2.0\,\text{L}$ , and $T_1=400\,\text{K}$ . It is changed to $P_2=75\,\text{kPa}$ and $V_2=4.0\,\text{L}$ while the amount of gas remains constant. Which statement correctly describes $T_2$ compared to $T_1$ ?

It is unchanged. (correct answer)
It is $200\,\text{K}$ larger.
It is twice as large.
It is half as large.

Explanation: This question tests the ideal gas law. For a fixed amount of gas, PV/T remains constant: P₁V₁/T₁ = P₂V₂/T₂. With P₁ = 150 kPa, V₁ = 2.0 L, T₁ = 400 K, P₂ = 75 kPa, and V₂ = 4.0 L, we solve for T₂: T₂ = T₁(P₂V₂)/(P₁V₁) = 400 K × (75 × 4)/(150 × 2) = 400 K × 300/300 = 400 K. Therefore T₂ = T₁, unchanged. Choice D incorrectly suggests adding 200 K, treating temperature as an additive quantity. Use the combined gas law P₁V₁/T₁ = P₂V₂/T₂ when multiple variables change.

Question 20

A piston-cylinder contains an ideal gas at $P_1=100\,\text{kPa}$ and $V_1=3.0\,\text{L}$ . The gas is compressed to $V_2=1.5\,\text{L}$ while temperature and moles are constant. Compared to $P_1$ , what is $P_2$ ?

It is $50\,\text{kPa}$ .
It is $200\,\text{kPa}$ . (correct answer)
It is $150\,\text{kPa}$ .
It is $100\,\text{kPa}$ because pressure does not depend on volume.

Explanation: This question tests the ideal gas law. For an ideal gas, PV = nRT relates pressure, volume, temperature, and amount of gas. When temperature and moles are constant, pressure and volume are inversely proportional: P₁V₁ = P₂V₂. With P₁ = 100 kPa, V₁ = 3.0 L, and V₂ = 1.5 L, we find P₂ = P₁V₁/V₂ = (100 kPa)(3.0 L)/(1.5 L) = 200 kPa. Choice D incorrectly assumes pressure is independent of volume, missing the inverse relationship. When temperature and moles are constant, use Boyle's Law: P₁V₁ = P₂V₂.