Question

1 mole of ideal gas is allowed to expand reversibly and adiabatically from a temperature of $27^{\circ} C$ The work done is $3\, kJ \,mol ^{-1}$ The final temperature of the gas is _______ $K$ (Nearest integer).
Given $C_{ V }=20 \,J\, mol ^{-1} K ^{-1}$

Answer 1

Correct Answer: 150

To find the final temperature of the gas after an adiabatic expansion, we use the formula for adiabatic processes:

\( W = C_V (T_i - T_f) \)

Where:

• \( W \) is the work done on/by the gas, \(3 \, \text{kJ} = 3000 \, \text{J} \).
• \( C_V \) is the molar heat capacity at constant volume, \(20 \, \text{J} \cdot \text{mol}^{-1} \cdot \text{K}^{-1} \).
• \( T_i \) is the initial temperature, \(T_i = 27^{\circ}C = 300 \, K \).
• \( T_f \) is the final temperature in Kelvin, which we need to find.

Rearrange the formula to solve for \( T_f \):

\( T_f = T_i - \frac{W}{C_V} \)

Substitute the known values:

\( T_f = 300 \, \text{K} - \frac{3000 \, \text{J}}{20 \, \text{J} \cdot \text{mol}^{-1} \cdot \text{K}^{-1}} \)

Simplify the calculation:

\( T_f = 300 \, \text{K} - 150 \, \text{K} \)

\( T_f = 150 \, \text{K} \)

The final temperature of the gas is 150 K, which falls within the expected range as specified.