An ideal gas ( = 1.5 ) initially at 1 litre, 300 K , and 100 kPa is suddenly compressed to half its original volume. Subsequently, the gas is cooled back to 300 K while maintaining its new constant pressure. Calculate the work done by the gas during this cooling phase.

Step-by-step solution

The gas is cooled at a constant pressure, making this an isobaric process. The work done in an isobaric process is W = nR(T₃ - T₂) . From the initial state, the gas constant factor can be evaluated as nR = P₁ V₁ T₁ = 100 300 = 1 3 J/K . The final temperature is T₃ = 300 K and the initial temperature for this phase is T₂ = 300 2 K . Substituting these values gives W = 1 3 (300 - 300 2 ) = 100(1 - 2 ) J . Using 2 1.414 , the work done is 100(1 - 1.414) = -41.4 J .