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.

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.

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.

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.

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.

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.

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.

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.

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.

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 $PV = nRT$.