Under which conditions does the molar volume of a gas decrease? at 273 K and at 2.0 atm at 273 K and at 0.5 atm at 409.5 K and at 1.0 atm at 546 K and at 2.0 atm

We start with the ideal gas law, which in molar form is \[ V_m = \frac{RT}{P}\,, \] where - \( V_m \) is the molar volume, - \( R \) is the gas constant, - \( T \) is the temperature (in Kelvin), and - \( P \) is the pressure. Notice that the molar volume is directly proportional to the temperature and inversely proportional to the pressure. This means that for a fixed amount of gas: - Lowering \( T \) will decrease \( V_m \), and - Increasing \( P \) will also decrease \( V_m \). Let’s analyze the four given conditions by comparing the ratio \(\frac{T}{P}\) (since \( V_m \propto \frac{T}{P} \)): 1. **At 273 K and 2.0 atm** \[ \frac{T}{P} = \frac{273}{2.0} = 136.5\,. \] This is a relatively small value, meaning the molar volume is quite low. 2. **At 273 K and 0.5 atm** \[ \frac{T}{P} = \frac{273}{0.5} = 546\,. \] A higher value than in (1), so the molar volume is larger. 3. **At 409.5 K and 1.0 atm** \[ \frac{T}{P} = \frac{409.5}{1.0} = 409.5\,. \] This is also larger than the value in condition (1). 4. **At 546 K and 2.0 atm** \[ \frac{T}{P} = \frac{546}{2.0} = 273\,. \] This value is in between those of conditions (1) and (3). Comparing these, the smallest ratio, and therefore the smallest molar volume, is under the condition with a low temperature and a high pressure, which is condition (1): **273 K and 2.0 atm**. Thus, the molar volume of a gas decreases under conditions where the temperature is lower and/or the pressure is higher. Among the given options, the one that produces the lowest molar volume is: - **At 273 K and at 2.0 atm.**