Question

11.0 L of an ideal gas at constant external pressure of 5 atm is compressed isothermally to a final volume of one liter. The heat absorbed and work done respectively, during this compression (in L atm) are

• A. -50, -50
• B. 50, -50
• C. -50, 50
• D. 50, 50

Hint:

In isothermal processes, the heat change is equal in magnitude and opposite in sign to work: \( q = -W \).

Answer 1

The Correct Option is C

Work done, \( W = -P_{\text{ext}} \Delta V = -5 (1 - 11) = -5 \times (-10) = 50\ \text{L atm} \)
Since the process is isothermal for an ideal gas: \( q = -W = -50\ \text{L atm} \)