Question

An ideal gas, $\overline{C_V} = \frac{5}{2} R$, is expanded adiabatically against a constant pressure of 1 atm until it doubles in volume.
If the initial temperature and pressure are $298 \, \text{K}$ and $5 \, \text{atm}$, respectively, then the final temperature is ______ $\, \text{K}$ (nearest integer).
Given: $\overline{C_V}$ is the molar heat capacity at constant volume.

Answer 1

For an adiabatic process:
$\Delta U = q + w, \quad q = 0 \implies \Delta U = w$
Step 1: Using the first law of thermodynamics:
$n C_V \Delta T = -P_{\text{ext}} (V_2 - V_1)$
Since $V_2 = 2V_1$, substitute and simplify:
$nR \frac{T_2}{P_2} = 2nR \frac{T_1}{P_1}.$
Step 2: Relation between $P_2$ and $T_2$:
$P_2 = \frac{5T_2}{2 \times 298}.$
Step 3: Using $C_V$:
$\frac{5}{2} nR (T_2 - T_1) = -nR T_1 \left( \frac{P_2}{P_1} - 1 \right).$
Step 4: Substitute and solve:
$T_2 = \frac{5T_2}{2 \times 298}.$
From the equation:
$T_2 \approx 274.16 \, \text{K}.$
Nearest integer:
$T_2 \approx 274 \, \text{K}.$

Answer 2

For an adiabatic process:
$\Delta U = q + w, \quad q = 0 \implies \Delta U = w$
Step 1: Using the first law of thermodynamics:
$n C_V \Delta T = -P_{\text{ext}} (V_2 - V_1)$
Since $V_2 = 2V_1$, substitute and simplify:
$nR \frac{T_2}{P_2} = 2nR \frac{T_1}{P_1}.$
Step 2: Relation between $P_2$ and $T_2$:
$P_2 = \frac{5T_2}{2 \times 298}.$
Step 3: Using $C_V$:
$\frac{5}{2} nR (T_2 - T_1) = -nR T_1 \left( \frac{P_2}{P_1} - 1 \right).$
Step 4: Substitute and solve:
$T_2 = \frac{5T_2}{2 \times 298}.$
From the equation:
$T_2 \approx 274.16 \, \text{K}.$
Nearest integer:
$T_2 \approx 274 \, \text{K}.$