the pressure of 1 mol of gas is decreased to 0.5 atm at 273 k. what happens to the molar volume of the gas…

Explanation:

Step1: Recall ideal gas law

$PV = nRT$. For 1 - mol gas ($n = 1$), $V=\frac{RT}{P}$ (where $R$ is the ideal - gas constant and $T$ is temperature).

Step2: Analyze the change in pressure

Initial pressure $P_1$ and volume $V_1$, final pressure $P_2$ and volume $V_2$. Since $T$ and $n$ are constant, $P_1V_1=P_2V_2$. Let the initial pressure $P_1 = 1$ atm (assume standard - like initial condition for comparison), and $P_2=0.5$ atm. Then $V_2=\frac{P_1}{P_2}V_1$. Substituting $P_1 = 1$ atm and $P_2 = 0.5$ atm, we get $V_2=\frac{1}{0.5}V_1 = 2V_1$.

Answer:

The molar volume is doubled.