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AP Chemistry

AP Chemistry Help: Ideal Gas Law

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

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}R=0.0821L⋅atm⋅mol−1⋅K−1.)

  1. 0.108 mol
  2. 0.0113 mol
  3. 1.33 mol
  4. 0.155 mol
  5. 0.133 mol
Explanation: This question tests the application of the ideal gas law, PV=nRTPV = nRTPV=nRT, to determine the amount of gas in moles. Use n=PVRTn = \frac{PV}{RT}n=RTPV​, 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}R=0.0821L⋅atm⋅mol−1⋅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}n=0.0821×2930.800×4.00​=0.133mol, 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 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 L0˘0b7atm0˘0b7mol−10˘0b7K−1R = 0.0821\ \text{L\u00b7atm\u00b7mol}^{-1}\text{\u00b7K}^{-1}R=0.0821 L0˘0b7atm0˘0b7mol−10˘0b7K−1.)

  1. 1.46 mol
  2. 0.0153 mol
  3. 2.44 mol
  4. 0.122 mol
  5. 0.183 mol
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

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 L0˘0b7atm0˘0b7mol−10˘0b7K−1R = 0.0821\ \text{L\u00b7atm\u00b7mol}^{-1}\text{\u00b7K}^{-1}R=0.0821 L0˘0b7atm0˘0b7mol−10˘0b7K−1.)

  1. 3.08 L
  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

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 L0˘0b7atm0˘0b7mol−10˘0b7K−1R = 0.0821\ \text{L\u00b7atm\u00b7mol}^{-1}\text{\u00b7K}^{-1}R=0.0821 L0˘0b7atm0˘0b7mol−10˘0b7K−1.)

  1. 0.229 mol
  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

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 L0˘0b7atm0˘0b7mol−10˘0b7K−1R = 0.0821\ \text{L\u00b7atm\u00b7mol}^{-1}\text{\u00b7K}^{-1}R=0.0821 L0˘0b7atm0˘0b7mol−10˘0b7K−1.)

  1. 1.29 L
  2. 129 L
  3. 12.9 L
  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

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

  1. 100 K100\,\text{K}100K
  2. 200 K200\,\text{K}200K
  3. 300 K300\,\text{K}300K
  4. 400 K400\,\text{K}400K
  5. 600 K600\,\text{K}600K
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}P=1.0atm, V=49.2 LV = 49.2 \, \text{L}V=49.2L, n=2.0 moln = 2.0 \, \text{mol}n=2.0mol, and R=0.082 L\cdotpatm\cdotpmol−1\cdotpK−1R = 0.082 \, \text{L·atm·mol}^{-1}\text{·K}^{-1}R=0.082L\cdotpatm\cdotpmol−1\cdotpK−1, we solve for TTT using T=PVnRT = \frac{PV}{nR}T=nRPV​. Substituting: T=1.0 atm×49.2 L2.0 mol×0.082 L\cdotpatm\cdotpmol−1\cdotpK−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·atm·mol}^{-1}\text{·K}^{-1}} = \frac{49.2}{0.164} = 300 \, \text{K}T=2.0mol×0.082L\cdotpatm\cdotpmol−1\cdotpK−11.0atm×49.2L​=0.16449.2​=300K. A common mistake is thinking you need to convert this to Celsius by subtracting 273, giving 27∘C27^\circ \text{C}27∘C, but the question asks for Kelvin. When using PV=nRTPV = nRTPV=nRT, the temperature calculated is always in Kelvin, matching the units of R.

Question 7

A balloon contains 0.25 mol0.25\,\text{mol}0.25mol of an ideal gas at a pressure of 1.0 atm1.0\,\text{atm}1.0atm and a volume of 6.15 L6.15\,\text{L}6.15L. What is the temperature of the gas in kelvins? (Use R=0.082 L\cdotpatm\cdotpmol−1\cdotpK−1R = 0.082\,\text{L·atm·mol}^{-1}\text{·K}^{-1}R=0.082L\cdotpatm\cdotpmol−1\cdotpK−1.)

  1. 30 K30\,\text{K}30K
  2. 300 K300\,\text{K}300K
  3. 273 K273\,\text{K}273K
  4. 615 K615\,\text{K}615K
  5. 200 K200\,\text{K}200K
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}P=1.0atm, V=6.15 LV = 6.15 \, \text{L}V=6.15L, n=0.25 moln = 0.25 \, \text{mol}n=0.25mol, and R=0.082 L\cdotpatm\cdotpmol−1\cdotpK−1R = 0.082 \, \text{L·atm·mol}^{-1}\text{·K}^{-1}R=0.082L\cdotpatm\cdotpmol−1\cdotpK−1, we solve for T using T=PVnRT = \frac{PV}{nR}T=nRPV​. Substituting: T=(1.0 atm×6.15 L)(0.25 mol×0.082 L\cdotpatm\cdotpmol−1\cdotpK−1)=6.150.0205=300 KT = \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}T=(0.25mol×0.082L\cdotpatm\cdotpmol−1\cdotpK−1)(1.0atm×6.15L)​=0.02056.15​=300K. 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

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

  1. 1.23 atm1.23\,\text{atm}1.23atm
  2. 12.3 atm12.3\,\text{atm}12.3atm
  3. 0.82 atm0.82\,\text{atm}0.82atm
  4. 0.12 atm0.12\,\text{atm}0.12atm
  5. 2.46 atm2.46\,\text{atm}2.46atm
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}n=0.50mol, V=10.0 LV = 10.0 \, \text{L}V=10.0L, T=27∘C=300 KT = 27^\circ \text{C} = 300 \, \text{K}T=27∘C=300K, and R=0.082 L\cdotpatm\cdotpmol−1\cdotpK−1R = 0.082 \, \text{L·atm·mol}^{-1}\text{·K}^{-1}R=0.082L\cdotpatm\cdotpmol−1\cdotpK−1, we solve for PPP using P=nRTVP = \frac{nRT}{V}P=VnRT​. Substituting: P=(0.50 mol)×(0.082 L\cdotpatm\cdotpmol−1\cdotpK−1)×(300 K)10.0 L=12.310.0=1.23 atmP = \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}P=10.0L(0.50mol)×(0.082L\cdotpatm\cdotpmol−1\cdotpK−1)×(300K)​=10.012.3​=1.23atm. A common mistake is using Celsius temperature directly (27∘C27^\circ \text{C}27∘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}P=10.0(0.50×0.082×27)​=0.11atm. Always convert temperature to Kelvin (K=∘C+273K = ^\circ \text{C} + 273K=∘C+273) before applying the ideal gas law.

Question 9

A 1.0 mol1.0\,\text{mol}1.0mol sample of an ideal gas is at 2.0 atm2.0\,\text{atm}2.0atm and 127∘C127^\circ\text{C}127∘C. What volume does the gas occupy? (Use R=0.082 L\cdotpatm\cdotpmol−1\cdotpK−1R = 0.082\,\text{L·atm·mol}^{-1}\text{·K}^{-1}R=0.082L\cdotpatm\cdotpmol−1\cdotpK−1.)

  1. 24.6 L24.6\,\text{L}24.6L
  2. 8.2 L8.2\,\text{L}8.2L
  3. 12.3 L12.3\,\text{L}12.3L
  4. 16.4 L16.4\,\text{L}16.4L
  5. 0.164 L0.164\,\text{L}0.164L
Explanation: This question tests your ability to calculate volume using the ideal gas law when given moles, pressure, and temperature. Given n=1.0 moln = 1.0 \, \text{mol}n=1.0mol, P=2.0 atmP = 2.0 \, \text{atm}P=2.0atm, T=127∘C=400 KT = 127^\circ \text{C} = 400 \, \text{K}T=127∘C=400K, and R=0.082 L\cdotpatm\cdotpmol−1\cdotpK−1R = 0.082 \, \text{L·atm·mol}^{-1}\text{·K}^{-1}R=0.082L\cdotpatm\cdotpmol−1\cdotpK−1, we solve for VVV using V=nRTPV = \frac{nRT}{P}V=PnRT​. Substituting: V=(1.0 mol)×(0.082 L\cdotpatm\cdotpmol−1\cdotpK−1)×(400 K)2.0 atm=32.82.0=16.4 LV = \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}V=2.0atm(1.0mol)×(0.082L\cdotpatm\cdotpmol−1\cdotpK−1)×(400K)​=2.032.8​=16.4L. A common error is forgetting to convert temperature to Kelvin, using 127 instead of 400, which would give V=(1.0)×(0.082)×(127)2.0=5.2 LV = \frac{(1.0) \times (0.082) \times (127)}{2.0} = 5.2 \, \text{L}V=2.0(1.0)×(0.082)×(127)​=5.2L. Remember to always add 273 to Celsius temperature to get Kelvin before using PV=nRTPV = nRTPV=nRT.

Question 10

A 0.10 mol0.10\,\text{mol}0.10mol sample of an ideal gas is in a 4.10 L4.10\,\text{L}4.10L container at 227∘C227^\circ\text{C}227∘C. What is the pressure of the gas? (Use R=0.082 L\cdotpatm\cdotpmol−1\cdotpK−1R = 0.082\,\text{L·atm·mol}^{-1}\text{·K}^{-1}R=0.082L\cdotpatm\cdotpmol−1\cdotpK−1.)

  1. 1.0 atm1.0\,\text{atm}1.0atm
  2. 0.10 atm0.10\,\text{atm}0.10atm
  3. 2.0 atm2.0\,\text{atm}2.0atm
  4. 0.50 atm0.50\,\text{atm}0.50atm
  5. 10 atm10\,\text{atm}10atm
Explanation: This question tests your ability to calculate pressure using the ideal gas law with given moles, volume, and temperature. With n=0.10 moln = 0.10 \, \text{mol}n=0.10mol, V=4.10 LV = 4.10 \, \text{L}V=4.10L, T=227∘C=500 KT = 227^\circ \text{C} = 500 \, \text{K}T=227∘C=500K, and R=0.082 L\cdotpatm\cdotpmol−1\cdotpK−1R = 0.082 \, \text{L·atm·mol}^{-1}\text{·K}^{-1}R=0.082L\cdotpatm\cdotpmol−1\cdotpK−1, we solve for PPP using P=nRTVP = \frac{nRT}{V}P=VnRT​. Substituting: P=(0.10 mol×0.082 L\cdotpatm\cdotpmol−1\cdotpK−1×500 K)4.10 L=4.14.10=1.0 atmP = \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}P=4.10L(0.10mol×0.082L\cdotpatm\cdotpmol−1\cdotpK−1×500K)​=4.104.1​=1.0atm. A common mistake is forgetting to convert Celsius to Kelvin, using 227 instead of 500, which would give P=(0.10×0.082×227)4.10=0.45 atmP = \frac{(0.10 \times 0.082 \times 227)}{4.10} = 0.45 \, \text{atm}P=4.10(0.10×0.082×227)​=0.45atm. Always add 273 to Celsius temperature to convert to Kelvin before using PV=nRTPV = nRTPV=nRT.