AP Chemistry Quiz: Ideal Gas Law

What this quiz covers

This quiz focuses on Ideal Gas Law, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Chemistry.

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.

Question 1

1. 0.108 mol
2. 0.0113 mol
3. 1.33 mol
4. 0.155 mol
5. 0.133 mol (correct answer)

Explanation: This question tests the application of the ideal gas law, $PV = nRT$, to determine the amount of gas in moles. Use $n = \frac{PV}{RT}$, with P = 0.800 atm, V = 4.00 L, T = 20°C converted to 293 K, and $R = 0.0821 \, \text{L} \cdot \text{atm} \cdot \text{mol}^{-1} \cdot \text{K}^{-1}$. Calculation yields $n = \frac{0.800 \times 4.00}{0.0821 \times 293} = 0.133 \, \text{mol}$, as per choice A. The law assumes ideal behavior where gases follow this relationship at moderate conditions. A tempting distractor is choice B, 0.0113 mol, which occurs if T = 20 K is used without conversion, highlighting the misconception of ignoring the Kelvin scale. A key strategy is to consistently convert temperatures to Kelvin and check if results make physical sense.

Question 2

1. 1.46 mol
2. 0.0153 mol
3. 2.44 mol
4. 0.122 mol
5. 0.183 mol (correct answer)

Explanation: This question tests the application of the ideal gas law, PV = nRT, to calculate the moles of gas in a container. Solve for n = PV/RT, converting T = 127°C to 400 K, with P = 1.20 atm, V = 5.00 L, and R = 0.0821 L·atm·mol⁻¹·K⁻¹. This gives n = (1.20 × 5.00) / (0.0821 × 400) = 0.183 mol, as in choice A. The ideal gas law relates these variables assuming negligible intermolecular forces and particle volume. A tempting distractor is choice E, 0.122 mol, which comes from using T = 127 K without conversion, embodying the misconception of failing to convert Celsius to Kelvin. Remember to always convert temperatures to Kelvin in gas law calculations to avoid errors in absolute temperature scales.

Question 3

1. 3.08 L (correct answer)
2. 30.8 L
3. 0.308 L
4. 4.10 L
5. 1.54 L

Explanation: This question tests the application of the ideal gas law, PV = nRT, to find the volume occupied by a gas sample. Rearrange to V = nRT/P, using n = 0.250 mol, T = 300 K, P = 2.00 atm, and R = 0.0821 L·atm·mol⁻¹·K⁻¹. Substituting provides V = (0.250 × 0.0821 × 300) / 2.00 = 3.08 L, matching choice A. This demonstrates volume's direct relation to moles and temperature, inverse to pressure. A tempting distractor is choice E, 1.54 L, resulting from dividing by 4.00 atm instead of 2.00 atm, due to the misconception of doubling the pressure value. To solve gas law problems effectively, verify all numerical values and perform calculations step by step.

Question 4

1. 0.229 mol (correct answer)
2. 0.305 mol
3. 2.29 mol
4. 0.0186 mol
5. 1.31 mol

Explanation: This question tests the application of the ideal gas law, PV = nRT, to calculate the moles of gas in a container. Use n = PV/RT, with P = 2.50 atm, V = 3.00 L, T = 400 K, and R = 0.0821 L·atm·mol⁻¹·K⁻¹. This yields n = (2.50 × 3.00) / (0.0821 × 400) = 0.229 mol, as in choice A. The equation holds for ideal gases where particles have negligible volume. A tempting distractor is choice D, 0.0186 mol, from using T = 40 K incorrectly, showing the misconception of not converting properly. A transferable strategy is to perform dimensional analysis to confirm units cancel correctly to the desired quantity.

Question 5

1. 1.29 L
2. 129 L
3. 12.9 L (correct answer)
4. 15.7 L
5. 0.0775 L

Explanation: This question tests the application of the ideal gas law, PV = nRT, to find the volume of a balloon containing gas. Rearrange to V = nRT/P, converting T = 25°C to 298 K, with n = 0.500 mol, P = 0.950 atm, and R = 0.0821 L·atm·mol⁻¹·K⁻¹. Substituting gives V = (0.500 × 0.0821 × 298) / 0.950 = 12.9 L, matching choice A. This illustrates volume's dependence on moles, temperature, and inverse pressure. A tempting distractor is choice B, 1.29 L, resulting from omitting the moles in the numerator, reflecting the misconception of forgetting a variable. When applying gas laws, list all known values and the target variable before calculating.

Question 6

1. $100\,\text{K}$
2. $200\,\text{K}$
3. $300\,\text{K}$ (correct answer)
4. $400\,\text{K}$
5. $600\,\text{K}$

Explanation: This question tests your ability to find temperature using the ideal gas law when given pressure, volume, and moles. With $P = 1.0 \, \text{atm}$, $V = 49.2 \, \text{L}$, $n = 2.0 \, \text{mol}$, and $R = 0.082 \, \text{L·atm·mol}^{-1}\text{·K}^{-1}$, we solve for $T$ using $T = \frac{PV}{nR}$. Substituting: $T = \frac{1.0 \, \text{atm} \times 49.2 \, \text{L}}{2.0 \, \text{mol} \times 0.082 \, \text{L·atm·mol}^{-1}\text{·K}^{-1}} = \frac{49.2}{0.164} = 300 \, \text{K}$. A common mistake is thinking you need to convert this to Celsius by subtracting 273, giving $27^\circ \text{C}$, but the question asks for Kelvin. When using $PV = nRT$, the temperature calculated is always in Kelvin, matching the units of R.

Question 7

1. $30\,\text{K}$
2. $300\,\text{K}$ (correct answer)
3. $273\,\text{K}$
4. $615\,\text{K}$
5. $200\,\text{K}$

Explanation: This question tests your ability to find temperature using the ideal gas law when given pressure, volume, and moles. With $P = 1.0 \, \text{atm}$, $V = 6.15 \, \text{L}$, $n = 0.25 \, \text{mol}$, and $R = 0.082 \, \text{L·atm·mol}^{-1}\text{·K}^{-1}$, we solve for T using $T = \frac{PV}{nR}$. Substituting: $T = \frac{(1.0 \, \text{atm} \times 6.15 \, \text{L})}{(0.25 \, \text{mol} \times 0.082 \, \text{L·atm·mol}^{-1}\text{·K}^{-1})} = \frac{6.15}{0.0205} = 300 \, \text{K}$. A common mistake is thinking the answer should be in Celsius and subtracting 273, which would give 27°C, but the question specifically asks for Kelvin. When the ideal gas law gives you temperature, it's always in Kelvin, so no conversion is needed if Kelvin is requested.

Question 8

1. $1.23\,\text{atm}$ (correct answer)
2. $12.3\,\text{atm}$
3. $0.82\,\text{atm}$
4. $0.12\,\text{atm}$
5. $2.46\,\text{atm}$

Explanation: This question tests your ability to use the ideal gas law to find pressure when given moles, volume, and temperature. With $n = 0.50 \, \text{mol}$, $V = 10.0 \, \text{L}$, $T = 27^\circ \text{C} = 300 \, \text{K}$, and $R = 0.082 \, \text{L·atm·mol}^{-1}\text{·K}^{-1}$, we solve for $P$ using $P = \frac{nRT}{V}$. Substituting: $P = \frac{(0.50 \, \text{mol}) \times (0.082 \, \text{L·atm·mol}^{-1}\text{·K}^{-1}) \times (300 \, \text{K})}{10.0 \, \text{L}} = \frac{12.3}{10.0} = 1.23 \, \text{atm}$. A common mistake is using Celsius temperature directly ($27^\circ \text{C}$) instead of converting to Kelvin, which would give $P = \frac{(0.50 \times 0.082 \times 27)}{10.0} = 0.11 \, \text{atm}$. Always convert temperature to Kelvin ($K = ^\circ \text{C} + 273$) before applying the ideal gas law.

Question 9

1. $24.6\,\text{L}$
2. $8.2\,\text{L}$
3. $12.3\,\text{L}$
4. $16.4\,\text{L}$ (correct answer)
5. $0.164\,\text{L}$

Explanation: This question tests your ability to calculate volume using the ideal gas law when given moles, pressure, and temperature. Given $n = 1.0 \, \text{mol}$, $P = 2.0 \, \text{atm}$, $T = 127^\circ \text{C} = 400 \, \text{K}$, and $R = 0.082 \, \text{L·atm·mol}^{-1}\text{·K}^{-1}$, we solve for $V$ using $V = \frac{nRT}{P}$. Substituting: $V = \frac{(1.0 \, \text{mol}) \times (0.082 \, \text{L·atm·mol}^{-1}\text{·K}^{-1}) \times (400 \, \text{K})}{2.0 \, \text{atm}} = \frac{32.8}{2.0} = 16.4 \, \text{L}$. A common error is forgetting to convert temperature to Kelvin, using 127 instead of 400, which would give $V = \frac{(1.0) \times (0.082) \times (127)}{2.0} = 5.2 \, \text{L}$. Remember to always add 273 to Celsius temperature to get Kelvin before using $PV = nRT$.

Question 10

1. $1.0\,\text{atm}$ (correct answer)
2. $0.10\,\text{atm}$
3. $2.0\,\text{atm}$
4. $0.50\,\text{atm}$
5. $10\,\text{atm}$

Explanation: This question tests your ability to calculate pressure using the ideal gas law with given moles, volume, and temperature. With $n = 0.10 \, \text{mol}$, $V = 4.10 \, \text{L}$, $T = 227^\circ \text{C} = 500 \, \text{K}$, and $R = 0.082 \, \text{L·atm·mol}^{-1}\text{·K}^{-1}$, we solve for $P$ using $P = \frac{nRT}{V}$. Substituting: $P = \frac{(0.10 \, \text{mol} \times 0.082 \, \text{L·atm·mol}^{-1}\text{·K}^{-1} \times 500 \, \text{K})}{4.10 \, \text{L}} = \frac{4.1}{4.10} = 1.0 \, \text{atm}$. A common mistake is forgetting to convert Celsius to Kelvin, using 227 instead of 500, which would give $P = \frac{(0.10 \times 0.082 \times 227)}{4.10} = 0.45 \, \text{atm}$. Always add 273 to Celsius temperature to convert to Kelvin before using $PV = nRT$.

Question 11

1. $0.020\,\text{mol}$
2. $0.10\,\text{mol}$ (correct answer)
3. $0.082\,\text{mol}$
4. $1.0\,\text{mol}$
5. $0.050\,\text{mol}$

Explanation: This question tests your ability to calculate moles using the ideal gas law when given pressure, volume, and temperature. With $P = 0.50 \, \text{atm}$, $V = 5.0 \, \text{L}$, $T = 27^\circ \text{C} = 300 \, \text{K}$, and $R = 0.082 \, \text{L·atm·mol}^{-1}\text{·K}^{-1}$, we solve for n using $n = \frac{PV}{RT}$. Substituting: $n = \frac{0.50 \, \text{atm} \times 5.0 \, \text{L}}{0.082 \, \text{L·atm·mol}^{-1}\text{·K}^{-1} \times 300 \, \text{K}} = \frac{2.5}{24.6} = 0.10 \, \text{mol}$. A common error is using Celsius temperature directly ($27^\circ \text{C}$) instead of Kelvin, which would give $n = \frac{2.5}{0.082 \times 27} = 1.13 \, \text{mol}$. Always remember to convert Celsius to Kelvin by adding 273 before applying the ideal gas law.

Question 12

1. 0.203 mol (correct answer)
2. 0.492 mol
3. 61.0 mol
4. 0.0167 mol
5. 0.0410 mol

Explanation: This question tests the application of the ideal gas law, PV = nRT, to determine the number of moles of gas. Rearrange the equation to n = PV/RT, with P = 0.500 atm, V = 10.0 L, T = 300 K, and R = 0.0821 L·atm·mol⁻¹·K⁻¹. Plugging in yields n = (0.500 × 10.0) / (0.0821 × 300) = 0.203 mol, corresponding to choice A. This uses the direct proportionality of moles to pressure and volume, and inverse to temperature, under ideal conditions. A tempting distractor is choice B, 0.492 mol, which arises from forgetting to include volume in the numerator, leading to the misconception of incomplete application of the formula. When solving ideal gas law problems, ensure all variables are correctly placed in the rearranged equation and units are consistent.

Question 13

1. 1.23 atm (correct answer)
2. 0.616 atm
3. 2.46 atm
4. 0.0554 atm
5. 12.3 atm

Explanation: This question tests the application of the ideal gas law, PV = nRT, to calculate the pressure of a gas sample. To find the pressure, rearrange the ideal gas law to P = nRT/V, using the given values of n = 0.100 mol, V = 2.00 L, T = 27°C (which must be converted to 300 K), and R = 0.0821 L·atm·mol⁻¹·K⁻¹. Substituting these values gives P = (0.100 × 0.0821 × 300) / 2.00 = 1.23 atm, matching choice A. The calculation relies on the ideal gas law assuming the gas behaves ideally under these conditions, with all units consistent with R. A tempting distractor is choice C, 2.46 atm, which results from mistakenly using V = 1.00 L instead of 2.00 L, reflecting a misconception of misreading the given volume. Always double-check unit conversions and given values before plugging into the ideal gas law equation.

Question 14

1. $150\,\text{K}$
2. $300\,\text{K}$ (correct answer)
3. $30.0\,\text{K}$
4. $600\,\text{K}$
5. $750\,\text{K}$

Explanation: This question tests applying the ideal gas law, PV = nRT, to find the temperature in Kelvin. Rearrange to T = PV / nR for computation. With P = 4.92 atm, V = 10.0 L, n = 2.00 mol, R = 0.0821 L·atm·mol⁻¹·K⁻¹. Substituting yields T = (4.92 × 10.0) / (2.00 × 0.0821) ≈ 300 K. A tempting distractor is 600 K, resulting from forgetting to divide by n = 2.00 mol, due to the misconception of treating n as 1. Always include the correct value for moles and verify the equation setup in gas law calculations.

Question 15

1. $0.146\,\text{mol}$ (correct answer)
2. $0.098\,\text{mol}$
3. $0.0120\,\text{mol}$
4. $1.46\,\text{mol}$
5. $0.876\,\text{mol}$

Explanation: This question tests the ideal gas law, PV = nRT, to calculate the moles of gas present. Solve for n = PV / RT with the given data. Using P = 1.20 atm, V = 3.00 L, T = 27°C or 300 K, R = 0.0821 L·atm·mol⁻¹·K⁻¹. This gives n = (1.20 × 3.00) / (0.0821 × 300) ≈ 0.146 mol. A tempting distractor is 0.098 mol, which occurs if temperature is not converted to Kelvin, reflecting the misconception of using Celsius directly. Develop the habit of temperature conversion and unit checking to excel in ideal gas law applications.

Question 16

1. $0.328\,\text{atm}$
2. $0.205\,\text{atm}$
3. $16.4\,\text{atm}$
4. $2.46\,\text{atm}$
5. $1.64\,\text{atm}$ (correct answer)

Explanation: This question tests the application of the ideal gas law, PV \= nRT, to determine the pressure of a gas sample. Rearrange the equation to solve for P \= \frac{nRT}{V}. Using n = 0.500 mol, T = 127°C converted to 400 K, V = 10.0 L, and R \= 0.0821 , \text{L·atm·mol}^{-1}\text{·K}^{-1}. Plugging in gives P \= \frac{0.500 \times 0.0821 \times 400}{10.0} \approx 1.64 , \text{atm}. A tempting distractor is 0.328 atm, which results from forgetting to convert temperature to Kelvin and using 127 K, embodying the misconception of using Celsius directly in the formula. To avoid errors, always convert temperatures to Kelvin and double-check the rearrangement of the ideal gas law.

Question 17

1. $0.821\,\text{atm}$
2. $1.64\,\text{atm}$ (correct answer)
3. $2.46\,\text{atm}$
4. $4.10\,\text{atm}$
5. $0.410\,\text{atm}$

Explanation: This question tests the ideal gas law, PV = nRT, to compute pressure in a rigid container. Solve for P = nRT / V using the given values. With n = 0.200 mol, T = 127°C or 400 K, V = 4.00 L, and R = 0.0821 L·atm·mol⁻¹·K⁻¹. This calculates to P = (0.200 × 0.0821 × 400) / 4.00 ≈ 1.64 atm. A tempting distractor is 0.821 atm, arising from using half the moles or forgetting to convert temperature, due to the misconception of incorrect temperature units. Consistently convert to Kelvin and verify all inputs to master ideal gas law problems.

Question 18

1. $12.3\,\text{L}$ (correct answer)
2. $6.15\,\text{L}$
3. $3.28\,\text{L}$
4. $0.615\,\text{L}$
5. $24.6\,\text{L}$

Explanation: This question tests the use of the ideal gas law, $PV = nRT$, to calculate the volume occupied by a gas. Solve for V by rearranging to $V = nRT / P$. The values are $n = 1.00 \, \text{mol}$, $T = 27^\circ \text{C}$ or $300 \, \text{K}$, $P = 2.00 \, \text{atm}$, and $R = 0.0821 \, \text{L·atm·mol}^{-1}\text{·K}^{-1}$. This yields $V = (1.00 \times 0.0821 \times 300) / 2.00 \approx 12.3 \, \text{L}$. A tempting distractor is $24.6 \, \text{L}$, obtained by neglecting to divide by $P = 2.00 \, \text{atm}$, due to the misconception of ignoring the pressure factor in the calculation. Remember to include all variables in the rearranged equation and confirm units match for accurate ideal gas law applications.

Question 19

1. 300 K (correct answer)
2. 27.0 K
3. 573 K
4. 246 K
5. 200 K

Explanation: This question tests the application of the ideal gas law, PV = nRT, to calculate the temperature of a gas. Solve for T = PV/nR, with P = 0.984 atm, V = 1.50 L, n = 0.0600 mol, and R = 0.0821 L·atm·mol⁻¹·K⁻¹. This results in T = (0.984 × 1.50) / (0.0600 × 0.0821) = 300 K, choice A. Temperature is directly proportional to pressure and volume, inversely to moles. A tempting distractor is choice D, 246 K, from using V = 1.00 L instead, due to the misconception of rounding or misreading volume. Always ensure accurate reading of all given data and use the correct rearranged form of the equation.

Question 20

1. 0.99 atm (correct answer)
2. 2.0 atm
3. 0.16 atm
4. 4.9 atm
5. 12 atm

Explanation: This question tests the application of the ideal gas law (PV = nRT) to calculate pressure. Given n = 0.20 mol, V = 5.0 L, T = 27°C = 300 K, and R = 0.082 L·atm·mol⁻¹·K⁻¹, we solve for pressure: P = nRT/V = (0.20 mol)(0.082 L·atm·mol⁻¹·K⁻¹)(300 K)/(5.0 L) = 4.92/5.0 = 0.984 atm ≈ 0.99 atm. The calculation confirms that choice A is correct. A common error (choice D, 4.9 atm) occurs when students forget to divide by the volume, getting P = nRT = 4.92 instead of P = nRT/V. When using PV = nRT, always identify which variable you're solving for and ensure all units are consistent before substituting values.