Question

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

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

Hint:

During compression, work done by gas is negative, so internal energy may increase more than the heat supplied.

Answer 1

The Correct Option is A

Concept:
According to the first law of thermodynamics: \[ \Delta U=Q-W \] where \(Q\) is heat supplied to the gas and \(W\) is work done by the gas.

Step 1: Given pressure: \[ P=50\,N/m^2 \]

Step 2: Initial volume: \[ V_1=10\,m^3 \] Final volume: \[ V_2=4\,m^3 \]

Step 3: Change in volume: \[ \Delta V=V_2-V_1=4-10=-6\,m^3 \]

Step 4: Work done by gas: \[ W=P\Delta V \] \[ W=50(-6) \] \[ W=-300\,J \]

Step 5: Heat added: \[ Q=100\,J \]

Step 6: First law: \[ \Delta U=Q-W \] \[ \Delta U=100-(-300) \] \[ \Delta U=400\,J \]

Step 7: Hence internal energy increases by: \[ \boxed{400\,J} \]