Question

Two moles of a gas at a temperature of \( 327^\circ\text{C} \) expands adiabatically such that its volume increases by 700%. If the ratio of the specific heat capacities of the gas is \( \frac{4}{3} \), then the work done by the gas is (Universal gas constant = \( 8.3 \, \text{Jmol}^{-1}\text{K}^{-1} \))

• A. 14.94 kJ
• B. 29.88 kJ
• C. 44.82 kJ
• D. 59.76 kJ

Hint:

"Increases by \( x% \)" means the final value is \( (1 + \frac{x}{100}) \) times the initial value. Here, \( V_2 = (1+7)V_1 = 8V_1 \).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
For an adiabatic process, the work done depends on the initial and final temperatures. We use the relation \( TV^{\gamma-1} = \text{constant} \) to find the final temperature.
Step 2: Key Formula or Approach:
1. Adiabatic Relation: \( T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1} \) 2. Work Done: \( W = \frac{nR(T_1 - T_2)}{\gamma - 1} \)
Step 3: Detailed Explanation:
Given: \( n = 2 \) moles \( T_1 = 327^\circ\text{C} = 327 + 273 = 600 \, \text{K} \) \( \gamma = \frac{4}{3} \) Volume increases by 700%, so \( V_2 = V_1 + 7V_1 = 8V_1 \). Calculate \( T_2 \): \[ T_1 V_1^{\frac{4}{3} - 1} = T_2 (8V_1)^{\frac{4}{3} - 1} \] \[ 600 \cdot V_1^{1/3} = T_2 \cdot (8V_1)^{1/3} \] \[ 600 = T_2 \cdot (8)^{1/3} \] \[ 600 = T_2 \cdot 2 \implies T_2 = 300 \, \text{K} \] Calculate Work Done \( W \): \[ W = \frac{nR(T_1 - T_2)}{\gamma - 1} \] \[ W = \frac{2 \times 8.3 \times (600 - 300)}{\frac{4}{3} - 1} \] \[ W = \frac{2 \times 8.3 \times 300}{1/3} \] \[ W = 2 \times 8.3 \times 300 \times 3 \] \[ W = 16.6 \times 900 = 14940 \, \text{J} \] \[ W = 14.94 \, \text{kJ} \]
Step 4: Final Answer:
The work done by the gas is 14.94 kJ.