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AP Chemistry Help: Ideal Gas Law

Review real example questions for Ideal Gas Law in AP Chemistry.

Question 1 / 10

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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}.)

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.