Ideal Gas Law - AP Chemistry

1. 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 $R = 0.082\,\text{L}\cdot\text{atm}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$.)​
   Answer: 0.98 atm 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.
2. A $0.25\ \text{mol}$ sample of an ideal gas is confined to a $5.0\ \text{L}$ container at $127^\circ\text{C}$. What is the pressure of the gas? (Use $R = 0.082\ \text{L}\cdot\text{atm}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$.)
   Answer: $1.64\ \text{atm}$ This question tests the ability to calculate pressure using the ideal gas law with a temperature conversion. Given n = 0.25 mol, V = 5.0 L, T = 127°C = 400 K, and R = 0.082 L·atm·mol⁻¹·K⁻¹, we solve for P. Using P = nRT/V = (0.25 mol)(0.082 L·atm·mol⁻¹·K⁻¹)(400 K)/(5.0 L) = 8.2/5.0 = 1.64 atm. Choice A (0.41 atm) might result from the misconception of using 127 K instead of converting 127°C to 400 K. Always convert Celsius to Kelvin by adding 273 before using the ideal gas law.
3. An ideal gas sample has a pressure of 1.00 atm and a volume of 24.6 L at 27C. How many moles of gas are present? (Use $R = 0.0821 \, \text{L} \cdot \text{atm} \cdot \text{mol}^{-1} \cdot \text{K}^{-1}$.)
   Answer: 1.00 mol This question tests the application of the ideal gas law, $PV = nRT$, to find the number of moles in a gas sample. Rearrange to $n = \frac{PV}{RT}$, converting T = 27°C to $300 \, \text{K}$, with P = 1.00 atm, V = 24.6 L, and $R = 0.0821 \, \text{L} \cdot \text{atm} \cdot \text{mol}^{-1} \cdot \text{K}^{-1}$. This provides $n = \frac{1.00 \times 24.6}{0.0821 \times 300} = 1.00 \, \text{mol}$, matching choice A. The result aligns with standard molar volume approximations at these conditions. A tempting distractor is choice E, 0.111 mol, resulting from using V = 2.46 L by decimal error, embodying the misconception of misinterpreting numerical values. To avoid common pitfalls, cross-check your calculations and ensure temperature is in Kelvin.
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 $R = 0.0821\ \text{L}\cdot\text{atm}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$.)
   Answer: 0.229 mol 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.
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 $R = 0.0821\ \text{L}\cdot\text{atm}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$.)
   Answer: 12.9 L 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.
6. A rigid $4.00\,\text{L}$ container holds $0.200\,\text{mol}$ of an ideal gas at $127^\circ\text{C}$. What is the pressure in atm? (Use $R = 0.0821\,\text{L}\cdot\text{atm}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$.)
   Answer: $1.64\,\text{atm}$ 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.
7. A $1.0\,\text{mol}$ sample of an ideal gas is at $2.0\,\text{atm}$ and $127^\circ\text{C}$. What volume does the gas occupy? (Use $R = 0.082\,\text{L}\cdot\text{atm}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$.)
   Answer: $16.4\,\text{L}$ 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 $R = 0.082 \, \text{L}\cdot\text{atm}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$, we solve for $V$ using $V = \frac{nRT}{P}$. Substituting: $V = \frac{(1.0 \, \text{mol}) \times (0.082 \, \text{L}\cdot\text{atm}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}) \times (400 \, \text{K})}{2.0 \, \text{atm}} = \frac{32.8}{2.0} = 16.4 \, \text{L}$. 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 $PV = nRT$.
8. A $2.00\,\text{mol}$ sample of an ideal gas is in a $10.0\,\text{L}$ container at a pressure of $4.92\,\text{atm}$. What is the temperature in K? (Use $R = 0.0821\,\text{L}\cdot\text{atm}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$.)
   Answer: $300\,\text{K}$ 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.
9. 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 $R = 0.0821\ \text{L}\cdot\text{atm}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$.)
   Answer: 3.08 L 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.
10. 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 $R = 0.0821\ \text{L}\cdot\text{atm}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$.)
    Answer: 0.183 mol 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.
11. A 2.0 L flask contains 0.10 mol of an ideal gas at 127°C. What is the pressure in atm? (Use $R = 0.082\,\text{L}\cdot\text{atm}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$.)
    Answer: 1.6 atm This question tests the application of the ideal gas law to calculate pressure with a temperature conversion. Given n = 0.10 mol, V = 2.0 L, T = 127°C = 400 K, and R = 0.082 L·atm·mol⁻¹·K⁻¹, we solve for pressure: P = nRT/V = (0.10 mol)(0.082 L·atm·mol⁻¹·K⁻¹)(400 K)/(2.0 L) = 3.28/2.0 = 1.64 atm ≈ 1.6 atm. This confirms choice C is correct. A common mistake (choice B, 0.52 atm) occurs when students forget to convert Celsius to Kelvin and use T = 127 instead of T = 400 K. Always convert temperature to Kelvin by adding 273 to the Celsius value before using the ideal gas law.
12. A $0.50\,\text{mol}$ sample of an ideal gas occupies $10.0\,\text{L}$ at $27^\circ\text{C}$. What is the pressure of the gas? (Use $R = 0.082\,\text{L}\cdot\text{atm}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$.)
    Answer: $1.23\,\text{atm}$ 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 $R = 0.082 \, \text{L}\cdot\text{atm}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$, we solve for $P$ using $P = \frac{nRT}{V}$. Substituting: $P = \frac{(0.50 \, \text{mol}) \times (0.082 \, \text{L}\cdot\text{atm}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}) \times (300 \, \text{K})}{10.0 \, \text{L}} = \frac{12.3}{10.0} = 1.23 \, \text{atm}$. A common mistake is using Celsius temperature directly ($27^\circ \text{C}$) 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.
13. A rigid 2.00 L flask contains 0.100 mol of an ideal gas at 27C. What is the pressure of the gas? (Use $R = 0.0821\ \text{L}\cdot\text{atm}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$.)
    Answer: 1.23 atm 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.
14. A $2.0\,\text{mol}$ sample of an ideal gas occupies $49.2\,\text{L}$ at $1.0\,\text{atm}$. What is the temperature of the gas in kelvins? (Use $R = 0.082\,\text{L}\cdot\text{atm}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$.)
    Answer: $300\,\text{K}$ 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 $R = 0.082 \, \text{L}\cdot\text{atm}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$, we solve for $T$ using $T = \frac{PV}{nR}$. Substituting: $T = \frac{1.0 \, \text{atm} \times 49.2 \, \text{L}}{2.0 \, \text{mol} \times 0.082 \, \text{L}\cdot\text{atm}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}} = \frac{49.2}{0.164} = 300 \, \text{K}$. A common mistake is thinking you need to convert this to Celsius by subtracting 273, giving $27^\circ \text{C}$, but the question asks for Kelvin. When using $PV = nRT$, the temperature calculated is always in Kelvin, matching the units of R.
15. A 0.500 mol sample of an ideal gas occupies 10.0 L at 127°C. What is the pressure of the gas in atm? (Use $R = 0.0821\ \text{L}\cdot\text{atm}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$.)
    Answer: $1.64\,\text{atm}$ 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 0.521 atm (choice A), 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.
16. 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 $R = 0.082\,\text{L}\cdot\text{atm}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$.)​
    Answer: 0.99 atm 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.
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 $R = 0.0821\ \text{L}\cdot\text{atm}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$.)
    Answer: 0.203 mol 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.
18. A $5.0\,\text{L}$ container holds an ideal gas at $27^\circ\text{C}$ and $0.50\,\text{atm}$. How many moles of gas are present? (Use $R = 0.082\,\text{L}\cdot\text{atm}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$.)
    Answer: $0.10\,\text{mol}$ 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 $R = 0.082 \, \text{L}\cdot\text{atm}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$, we solve for n using $n = \frac{PV}{RT}$. Substituting: $n = \frac{0.50 \, \text{atm} \times 5.0 \, \text{L}}{0.082 \, \text{L}\cdot\text{atm}\cdot\text{mol}^{-1}\cdot\text{K}^{-1} \times 300 \, \text{K}} = \frac{2.5}{24.6} = 0.10 \, \text{mol}$. 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.
19. A container holds $1.00\,\text{mol}$ of an ideal gas at $2.00\,\text{atm}$ and $27^\circ\text{C}$. What volume does the gas occupy? (Use $R = 0.0821\,\text{L}\cdot\text{atm}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$.)
    Answer: $12.3\,\text{L}$ 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 $R = 0.0821 \, \text{L}\cdot\text{atm}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$. 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.
20. A gas sample in a $1.50\,\text{L}$ flask has a pressure of $0.800\,\text{atm}$ and a temperature of $77^\circ\text{C}$. Assuming ideal behavior, how many moles of gas are in the flask? (Use $R = 0.0821\,\text{L}\cdot\text{atm}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$.)
    Answer: $0.042\,\text{mol}$ 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.

Start this quiz