Question

A gas in a cylinder is compressed from an initial volume of \( 5 \, \text{m}^3 \) to a final volume of \( 2 \, \text{m}^3 \) while maintaining a constant pressure of \( 1 \times 10^5 \, \text{Pa} \). Calculate the work done by the gas during the compression.

• A. \( -3 \times 10^5 \, \text{J} \)
• B. \( -1 \times 10^5 \, \text{J} \)
• C. \( 3 \times 10^5 \, \text{J} \)
• D. \( 1 \times 10^5 \, \text{J} \)

Hint:

In an isobaric process, when a gas is compressed, the work done by the gas is negative. The negative sign indicates that energy is transferred to the surroundings.

Answer 1

The Correct Option is A

The work done by a gas during an isobaric process (constant pressure) is given by the formula: \[ W = - P \Delta V \] Where: - \( W \) is the work done by the gas, - \( P = 1 \times 10^5 \, \text{Pa} \) is the pressure, - \( \Delta V = V_f - V_i = 2 - 5 = -3 \, \text{m}^3 \) is the change in volume. Thus, the work done is: \[ W = - (1 \times 10^5) \times (-3) = 3 \times 10^5 \, \text{J} \] The negative sign indicates that the gas is being compressed, meaning the work done on the gas is negative. Therefore, the work done by the gas is \( -3 \times 10^5 \, \text{J} \).