An ideal gas system changes from state i to state f by paths iaf and ibf. If the heat added along iaf is Q_iaf = 50 cal, the work along iaf is W_iaf = 20 cal. Along ibf, if Q_ibf = 40 cal, the work done, W_iaf, is A) 10 cal. B) 20 cal. C) 30 cal. D) 40 cal. E) 50 cal.

Question

Quizplus · Accepted Answer

The first law of thermodynamics states that ΔU = Q - W, where ΔU is the change in internal energy of the system, Q is the heat added to the system, and W is the work done by the system. 
For the path iaf, we know that Q = 50 cal and W = 20 cal. Therefore, ΔUiaf = 50 - 20 = 30 cal. 
For the entire process (paths iaf and ibf), we know that ΔU = 0 (since the system starts and ends at the same state). Therefore, ΔUibf = -ΔUiaf = -30 cal. 
Also, we know that Qibf = 40 cal. Therefore, Wibf = Qibf - ΔUibf = 40 - (-30) = 70 cal. 
The work done along ibf is therefore 70 - 20 = 50 cal.

An Ideal Gas System Changes from State I to State