Question

During adiabatic expansion, if the temperature of 3 moles of a diatomic gas decreases by \(50^\circ \text{C}\), then the work done by the gas is
(R - Universal gas constant)

• A. \(375R\)
• B. \(750R\)
• C. \(1500R\)
• D. \(825R\)

Hint:

For adiabatic processes, work is calculated using \( W = \frac{nR\Delta T}{\gamma - 1} \). Always convert temperature changes to Kelvin and note the sign.

Answer 1

The Correct Option is A

Step 1: Use the formula for work done in adiabatic expansion of a diatomic gas: \[ W = \frac{n R \Delta T}{\gamma - 1} \] For a diatomic gas, \( \gamma = \frac{7}{5} \Rightarrow \gamma - 1 = \frac{2}{5} \) 
Step 2: Given: \( n = 3 \), \( \Delta T = -50^\circ\text{C} = -50\,\text{K} \) \[ W = \frac{3 R (-50)}{2/5} = -75R \cdot \frac{5}{2} = -187.5R \] The work done by the gas is positive during expansion, so magnitude is: \[ W = 187.5R \times 2 = 375R \]