Question

Initially the pressure of 1 mole of an ideal gas is \( 10^5 \) Nm\(^2\) and its volume is 16 liters. When it is adiabatically compressed, its final volume is 2 liters. Work done on the gas is (molar specific heat at constant volume \( C_V = \frac{3R}{2} \)):

• A. \( 72 \) kJ
• B. \( 7.2 \) kJ
• C. \( 720 \) kJ
• D. \( 360 \) kJ

Hint:

For an adiabatic process, use the relation \( W = \frac{P_1 V_1 - P_2 V_2}{\gamma - 1} \). Ensure volume is in cubic meters before calculation.

Answer 1

The Correct Option is B

Step 1: Recall the formula for work done in an adiabatic process The work done in an adiabatic process is given by: \[ W = \frac{P_1 V_1 - P_2 V_2}{\gamma - 1} \] Where: - \(P_1 = 10^5 \text{ N/m}^2\) - \(V_1 = 16 \text{ L} = 16 \times 10^{-3} \text{ m}^3\) - \(V_2 = 2 \text{ L} = 2 \times 10^{-3} \text{ m}^3\) The ratio of specific heats is: \[ \gamma = \frac{C_p}{C_v} = \frac{5R/2}{3R/2} = \frac{5}{3} \]
Step 2: Finding the final pressure using the adiabatic condition From the adiabatic relation: \[ P_1 V_1^\gamma = P_2 V_2^\gamma \] \[ P_2 = P_1 \left( \frac{V_1}{V_2} \right)^\gamma \] \[ P_2 = 10^5 \left( \frac{16}{2} \right)^{5/3} \] \[ P_2 = 10^5 \times 8^{5/3} \] Since \( 8^{5/3} \approx 32 \), \[ P_2 \approx 32 \times 10^5 = 3.2 \times 10^6 \text{ N/m}^2 \]
Step 3: Calculate the Work Done \[ W = \frac{P_1 V_1 - P_2 V_2}{\gamma - 1} \] \[ W = \frac{(10^5)(16 \times 10^{-3}) - (3.2 \times 10^6)(2 \times 10^{-3})}{\frac{5}{3} - 1} \] \[ W = \frac{(1600) - (6400)}{\frac{5}{3} - 1} \] \[ W = \frac{-4800}{\frac{2}{3}} = -4800 \times \frac{3}{2} = -7200 \text{ J} = -7.2 \text{ kJ} \] (Note: The negative sign indicates work is done *on* the gas.)