Question

A perfect gas goes from state A to state B by absorbing 8 × 10⁵ J of heat and doing 6.5 × 10⁵ J of external work. It is now transferred between the same two states in another process in which it absorbs 10s J of heat. In second process. Find the work done in the second process.

Answer 1

According to the first law of thermodynamics

∆U = ∆Q – ∆W

In the first process ∆U = 8 × 10⁵ – 6.5 × 10⁵ = 1.5 × 10⁵ J

Now ∆U, being a state function, remains the same in the second process,

∆W = ∆Q – ∆U = 1 × 10⁵ – 1.5 × 10⁵ ∆W = – 0.5 × 10⁵ J 

The negative sign shows that work is done on the gas.