Question

A gas undergoes a process \( PV^2 = \text{constant} \). Initially, pressure = 2 atm, volume = 1 L. It expands to 2 L. Find the work done by the gas in joules.

• A. 50.5 J
• B. 10.1 J
• C. 202 J
• D. 303 J

Hint:

For processes where \( PV^n = \text{constant} \), use the formula \( W = \int P \, dV \) to calculate the work done by the gas.

Answer 1

The Correct Option is B

For the given process \( PV^2 = \text{constant} \), the work done by the gas during the expansion is given by the integral: \[ W = \int_{V_1}^{V_2} P \, dV \] Since \( P = \frac{\text{constant}}{V^2} \), the work done can be calculated as: \[ W = \int_{V_1}^{V_2} \frac{C}{V^2} dV \] Using the limits \( V_1 = 1 \, \text{L} \) and \( V_2 = 2 \, \text{L} \), and knowing the relationship \( P_1 V_1^2 = P_2 V_2^2 \), we can calculate the work done. The pressure at \( V_2 \) can be found from the initial condition: \[ P_1 V_1^2 = P_2 V_2^2 \] Thus, the work done by the gas is approximately 10.1 J.