AP Chemistry Quiz: Ideal Gas Law
20 questions · exam conditions
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Ideal Gas LawQuestion 1 of 20

An ideal gas sample occupies 4.00 L at 0.800 atm and 20C. What is the amount of gas present? (Use R=0.0821 L⋅atm⋅mol−1⋅K−1R = 0.0821 \, \text{L} \cdot \text{atm} \cdot \text{mol}^{-1} \cdot \text{K}^{-1}.)

0.108 mol
0.0113 mol
1.33 mol
0.155 mol
0.133 mol
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AP Chemistry Quiz

AP Chemistry Quiz: Ideal Gas Law

Practice Ideal Gas Law in AP Chemistry 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 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.

All questions

Question 1

An ideal gas sample occupies 4.00 L at 0.800 atm and 20C. What is the amount of gas present? (Use R=0.0821 L⋅atm⋅mol−1⋅K−1R = 0.0821 \, \text{L} \cdot \text{atm} \cdot \text{mol}^{-1} \cdot \text{K}^{-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=nRTPV = nRT, to determine the amount of gas in moles. Use n=PVRTn = \frac{PV}{RT}, with P = 0.800 atm, V = 4.00 L, T = 20°C converted to 293 K, and R=0.0821 L⋅atm⋅mol−1⋅K−1R = 0.0821 \, \text{L} \cdot \text{atm} \cdot \text{mol}^{-1} \cdot \text{K}^{-1}. Calculation yields n=0.800×4.000.0821×293=0.133 moln = \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

A balloon contains 0.500 mol of an ideal gas at 25C and a pressure of 0.950 atm. What is the volume of the balloon? (Use R=0.0821 L⋅atm⋅mol−1⋅K−1R = 0.0821\ \text{L}\cdot\text{atm}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}.)

  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 3

A 2.00 mol2.00\,\text{mol} sample of an ideal gas is in a 10.0 L10.0\,\text{L} container at a pressure of 4.92 atm4.92\,\text{atm}. What is the temperature in K? (Use R=0.0821 L⋅atm⋅mol−1⋅K−1R = 0.0821\,\text{L}\cdot\text{atm}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}.)

  1. 150 K150\,\text{K}
  2. 300 K300\,\text{K} (correct answer)
  3. 30.0 K30.0\,\text{K}
  4. 600 K600\,\text{K}
  5. 750 K750\,\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 4

A student has 0.0400 mol of an ideal gas in a 1.00 L flask at 300 K. What pressure (in atm) does the gas exert? (Use R=0.0821 L⋅atm⋅mol−1⋅K−1R = 0.0821\ \text{L}\cdot\text{atm}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}.)

  1. 0.985 atm (correct answer)
  2. 9.85 atm
  3. 0.109 atm
  4. 1.31 atm
  5. 0.00328 atm
Explanation: This question tests the application of the ideal gas law, PV = nRT, to determine the pressure exerted by a gas. Solve for P = nRT/V, with n = 0.0400 mol, T = 300 K, V = 1.00 L, and R = 0.0821 L·atm·mol⁻¹·K⁻¹. Calculation gives P = (0.0400 × 0.0821 × 300) / 1.00 = 0.985 atm, choice A. Pressure relates directly to moles and temperature, inversely to volume. A tempting distractor is choice B, 9.85 atm, from using n = 0.400 mol by misplacing the decimal, due to the misconception of reading errors in quantities. Always verify calculations with approximate values to check if the answer is reasonable.

Question 5

A 0.300 mol0.300\,\text{mol} sample of an ideal gas is at 47∘C47^\circ\text{C} and 2.00 atm2.00\,\text{atm}. What volume does the gas occupy? (Use R=0.0821 L⋅atm⋅mol−1⋅K−1R = 0.0821\,\text{L}\cdot\text{atm}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}.)

  1. 3.94 L3.94\,\text{L} (correct answer)
  2. 0.985 L0.985\,\text{L}
  3. 7.88 L7.88\,\text{L}
  4. 1.97 L1.97\,\text{L}
  5. 4.80 L4.80\,\text{L}
Explanation: This question tests using the ideal gas law, PV = nRT, to calculate the volume occupied by a gas sample. Solve for V = nRT / P with the provided values. Using n = 0.300 mol, T = 47°C or 320 K, P = 2.00 atm, R = 0.0821 L·atm·mol⁻¹·K⁻¹. The result is V = (0.300 × 0.0821 × 320) / 2.00 ≈ 3.94 L. A tempting distractor is 1.97 L, which comes from dividing by P twice, reflecting the misconception of repeating the pressure term in the denominator. To prevent such errors, write out the formula clearly and check each step in ideal gas law solutions.

Question 6

A gas sample in a 1.50 L1.50\,\text{L} flask has a pressure of 0.800 atm0.800\,\text{atm} and a temperature of 77∘C77^\circ\text{C}. Assuming ideal behavior, how many moles of gas are in the flask? (Use R=0.0821 L⋅atm⋅mol−1⋅K−1R = 0.0821\,\text{L}\cdot\text{atm}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}.)

  1. 0.146 mol0.146\,\text{mol}
  2. 0.084 mol0.084\,\text{mol}
  3. 0.0042 mol0.0042\,\text{mol}
  4. 0.021 mol0.021\,\text{mol}
  5. 0.042 mol0.042\,\text{mol} (correct answer)
Explanation: This question tests the ideal gas law, PV = nRT, to determine the moles in a flask assuming ideal behavior. Use n = PV / RT for the calculation. Given P = 0.800 atm, V = 1.50 L, T = 77°C converted to 350 K, R = 0.0821 L·atm·mol⁻¹·K⁻¹. This computes to n = (0.800 × 1.50) / (0.0821 × 350) ≈ 0.042 mol. A tempting distractor is 0.021 mol, arising from halving the pressure or volume incorrectly, embodying a misconception in multiplying PV. Practice step-by-step substitution and arithmetic verification to handle ideal gas law problems effectively.

Question 7

A 0.50 mol0.50\,\text{mol} sample of an ideal gas occupies 10.0 L10.0\,\text{L} at 27∘C27^\circ\text{C}. What is the pressure of the gas? (Use R=0.082 L⋅atm⋅mol−1⋅K−1R = 0.082\,\text{L}\cdot\text{atm}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}.)

  1. 1.23 atm1.23\,\text{atm} (correct answer)
  2. 12.3 atm12.3\,\text{atm}
  3. 0.82 atm0.82\,\text{atm}
  4. 0.12 atm0.12\,\text{atm}
  5. 2.46 atm2.46\,\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 moln = 0.50 \, \text{mol}, V=10.0 LV = 10.0 \, \text{L}, T=27∘C=300 KT = 27^\circ \text{C} = 300 \, \text{K}, and R=0.082 L⋅atm⋅mol−1⋅K−1R = 0.082 \, \text{L}\cdot\text{atm}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}, we solve for PP using P=nRTVP = \frac{nRT}{V}. Substituting: P=(0.50 mol)×(0.082 L⋅atm⋅mol−1⋅K−1)×(300 K)10.0 L=12.310.0=1.23 atmP = \frac{(0.50 \, \text{mol}) \times (0.082 \, \text{L}\cdot\text{atm}\cdot\text{mol}^{-1}\cdot\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∘C27^\circ \text{C}) instead of converting to Kelvin, which would give P=(0.50×0.082×27)10.0=0.11 atmP = \frac{(0.50 \times 0.082 \times 27)}{10.0} = 0.11 \, \text{atm}. Always convert temperature to Kelvin (K=∘C+273K = ^\circ \text{C} + 273) before applying the ideal gas law.

Question 8

A 3.00 L container holds an ideal gas at 2.50 atm and 400 K. What is the amount of gas in the container? (Use R=0.0821 L⋅atm⋅mol−1⋅K−1R = 0.0821\ \text{L}\cdot\text{atm}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}.)

  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 9

A 0.50 mol0.50\ \text{mol} sample of an ideal gas exerts a pressure of 2.00 atm2.00\ \text{atm} at 300 K300\ \text{K}. What volume does it occupy? (Use R=0.082 L⋅atm⋅mol−1⋅K−1R = 0.082\ \text{L}\cdot\text{atm}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}.)

  1. 6.15 L6.15\ \text{L} (correct answer)
  2. 12.3 L12.3\ \text{L}
  3. 3.08 L3.08\ \text{L}
  4. 24.6 L24.6\ \text{L}
  5. 0.82 L0.82\ \text{L}
Explanation: This question tests the ability to calculate volume using the ideal gas law. Given n = 0.50 mol, P = 2.00 atm, T = 300 K, and R = 0.082 L·atm·mol⁻¹·K⁻¹, we solve for V. Rearranging PV = nRT gives V = nRT/P = (0.50 mol)(0.082 L·atm·mol⁻¹·K⁻¹)(300 K)/(2.00 atm) = 12.3/2.00 = 6.15 L. Choice B (12.3 L) represents the misconception of forgetting to divide by pressure, calculating only nRT. When solving for any variable in the ideal gas law, ensure you properly rearrange the equation before substituting values.

Question 10

A sample of an ideal gas has a pressure of 2.0 atm and occupies 3.0 L at 300 K. What amount of gas, in moles, is present? (Use R=0.082 L⋅atm⋅mol−1⋅K−1R = 0.082\,\text{L}\cdot\text{atm}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}.)​

  1. 0.082 mol
  2. 0.24 mol (correct answer)
  3. 0.020 mol
  4. 24 mol
  5. 1.6 mol
Explanation: This question tests the application of the ideal gas law to calculate the amount of gas in moles. Given P = 2.0 atm, V = 3.0 L, T = 300 K, and R = 0.082 L·atm·mol⁻¹·K⁻¹, we solve for n: n = PV/RT = (2.0 atm)(3.0 L)/[(0.082 L·atm·mol⁻¹·K⁻¹)(300 K)] = 6.0/24.6 = 0.244 mol ≈ 0.24 mol. This confirms choice B is correct. A common mistake (choice C, 0.020 mol) results from incorrectly multiplying P × V × R × T instead of dividing PV by RT, showing confusion about algebraic manipulation. To avoid errors, first rearrange PV = nRT algebraically to isolate your unknown variable, then substitute values with units to check dimensional consistency.

Question 11

A rigid 5.0 L container holds 0.20 mol of an ideal gas at 27°C. Assuming ideal behavior, what is the pressure of the gas in atm? (Use R=0.082 L⋅atm⋅mol−1⋅K−1R = 0.082\,\text{L}\cdot\text{atm}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}.)​

  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.

Question 12

A rigid 2.00 L flask contains 0.100 mol of an ideal gas at 27C. What is the pressure of the gas? (Use R=0.0821 L⋅atm⋅mol−1⋅K−1R = 0.0821\ \text{L}\cdot\text{atm}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}.)

  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 13

An ideal gas sample has a pressure of 1.20 atm1.20\,\text{atm}, a volume of 3.00 L3.00\,\text{L}, and a temperature of 27∘C27^\circ\text{C}. How many moles of gas are present? (Use R=0.0821 L⋅atm⋅mol−1⋅K−1R = 0.0821\,\text{L}\cdot\text{atm}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}.)

  1. 0.146 mol0.146\,\text{mol} (correct answer)
  2. 0.098 mol0.098\,\text{mol}
  3. 0.0120 mol0.0120\,\text{mol}
  4. 1.46 mol1.46\,\text{mol}
  5. 0.876 mol0.876\,\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 14

A student collects 0.040 mol of an ideal gas in a 1.0 L container at 27°C. What pressure should the gas exert in atm? (Use R=0.082 L⋅atm⋅mol−1⋅K−1R = 0.082\,\text{L}\cdot\text{atm}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}.)​

  1. 0.33 atm
  2. 0.98 atm (correct answer)
  3. 1.2 atm
  4. 9.8 atm
  5. 33 atm
Explanation: This question tests the application of the ideal gas law to calculate pressure. Given n = 0.040 mol, V = 1.0 L, T = 27°C = 300 K, and R = 0.082 L·atm·mol⁻¹·K⁻¹, we solve for pressure: P = nRT/V = (0.040 mol)(0.082 L·atm·mol⁻¹·K⁻¹)(300 K)/(1.0 L) = 0.984/1.0 = 0.984 atm ≈ 0.98 atm. This confirms choice B is correct. A common mistake (choice A, 0.33 atm) might result from using temperature in Celsius (27) instead of Kelvin (300), giving P = 0.033 × 27 ≈ 0.89 atm, or other calculation errors. Remember to always convert temperature to Kelvin and check that your answer has reasonable magnitude for the given conditions.

Question 15

A student collects an ideal gas in a 5.00 L container at 1.20 atm and 127C. What amount of gas (in mol) is in the container? (Use R=0.0821 L⋅atm⋅mol−1⋅K−1R = 0.0821\ \text{L}\cdot\text{atm}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}.)

  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 16

A sealed 1.50 L container holds 0.0600 mol of an ideal gas at a pressure of 0.984 atm. What is the temperature of the gas in Kelvin? (Use R=0.0821 L⋅atm⋅mol−1⋅K−1R = 0.0821\ \text{L}\cdot\text{atm}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}.)

  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 17

A sample of an ideal gas has a pressure of 0.500 atm and occupies 10.0 L at 300 K. How many moles of gas are present? (Use R=0.0821 L⋅atm⋅mol−1⋅K−1R = 0.0821\ \text{L}\cdot\text{atm}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}.)

  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 18

A 0.250 mol sample of an ideal gas exerts a pressure of 2.00 atm at 300 K. What volume does the gas occupy? (Use R=0.0821 L⋅atm⋅mol−1⋅K−1R = 0.0821\ \text{L}\cdot\text{atm}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}.)

  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 19

A container holds 1.00 mol1.00\,\text{mol} of an ideal gas at 2.00 atm2.00\,\text{atm} and 27∘C27^\circ\text{C}. What volume does the gas occupy? (Use R=0.0821 L⋅atm⋅mol−1⋅K−1R = 0.0821\,\text{L}\cdot\text{atm}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}.)

  1. 12.3 L12.3\,\text{L} (correct answer)
  2. 6.15 L6.15\,\text{L}
  3. 3.28 L3.28\,\text{L}
  4. 0.615 L0.615\,\text{L}
  5. 24.6 L24.6\,\text{L}
Explanation: This question tests the use of the ideal gas law, PV=nRTPV = nRT, to calculate the volume occupied by a gas. Solve for V by rearranging to V=nRT/PV = nRT / P. The values are n=1.00 moln = 1.00 \, \text{mol}, T=27∘CT = 27^\circ \text{C} or 300 K300 \, \text{K}, P=2.00 atmP = 2.00 \, \text{atm}, and R=0.0821 L⋅atm⋅mol−1⋅K−1R = 0.0821 \, \text{L}\cdot\text{atm}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}. This yields V=(1.00×0.0821×300)/2.00≈12.3 LV = (1.00 \times 0.0821 \times 300) / 2.00 \approx 12.3 \, \text{L}. A tempting distractor is 24.6 L24.6 \, \text{L}, obtained by neglecting to divide by P=2.00 atmP = 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 20

A 2.0 mol2.0\,\text{mol} sample of an ideal gas occupies 49.2 L49.2\,\text{L} at 1.0 atm1.0\,\text{atm}. What is the temperature of the gas in kelvins? (Use R=0.082 L⋅atm⋅mol−1⋅K−1R = 0.082\,\text{L}\cdot\text{atm}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}.)

  1. 100 K100\,\text{K}
  2. 200 K200\,\text{K}
  3. 300 K300\,\text{K} (correct answer)
  4. 400 K400\,\text{K}
  5. 600 K600\,\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 atmP = 1.0 \, \text{atm}, V=49.2 LV = 49.2 \, \text{L}, n=2.0 moln = 2.0 \, \text{mol}, and R=0.082 L⋅atm⋅mol−1⋅K−1R = 0.082 \, \text{L}\cdot\text{atm}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}, we solve for TT using T=PVnRT = \frac{PV}{nR}. Substituting: T=1.0 atm×49.2 L2.0 mol×0.082 L⋅atm⋅mol−1⋅K−1=49.20.164=300 KT = \frac{1.0 \, \text{atm} \times 49.2 \, \text{L}}{2.0 \, \text{mol} \times 0.082 \, \text{L}\cdot\text{atm}\cdot\text{mol}^{-1}\cdot\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∘C27^\circ \text{C}, but the question asks for Kelvin. When using PV=nRTPV = nRT, the temperature calculated is always in Kelvin, matching the units of R.