Question

A gas is compressed at a constant pressure of \(50\text{ N/m}^2\) from a volume of \(10\text{ m}^3\) to a volume of \(4\text{ m}^3\). Energy of \(100\text{ J}\) is then added to the gas by heating. Its internal energy is

• A. Increases by \(400\text{ J}\)
• B. Increases by \(200\text{ J}\)
• C. Increases by \(100\text{ J}\)
• D. Decreases by \(200\text{ J}\)

Hint:

Use first law \(Q=\Delta U+W\), where \(W=P\Delta V\) is work done by the gas. Compression gives negative \(W\).

Answer 1

The Correct Option is A

Given pressure: \[ P=50\text{ N/m}^2. \] Initial volume: \[ V_1=10\text{ m}^3. \] Final volume: \[ V_2=4\text{ m}^3. \] Heat added to the gas: \[ Q=100\text{ J}. \] Change in volume is: \[ \Delta V=V_2-V_1. \] \[ \Delta V=4-10=-6\text{ m}^3. \] Work done by gas is: \[ W=P\Delta V. \] \[ W=50(-6). \] \[ W=-300\text{ J}. \] Negative work means work is done on the gas. Using first law of thermodynamics: \[ Q=\Delta U+W. \] Therefore: \[ \Delta U=Q-W. \] Substitute: \[ \Delta U=100-(-300). \] \[ \Delta U=100+300. \] \[ \Delta U=400\text{ J}. \] Thus, internal energy increases by: \[ 400\text{ J}. \]