Process one no work is done as volume in constant. However process 2 work is done as volume is not constant. Why is there no mention of this in the mark scheme?
Thermodynamics (cycles) Watch
- Thread Starter
- 18-01-2015 12:11
- 18-01-2015 13:06
The mark scheme DOES state that in process 1 for part a. the work done is zero because the change in volume is zero. (Line 2 under the two graphs.
"W = 0 (as ΔV = 0)"
As the second process in part a. is isothermal ΔU is zero. The question is asking you to find ΔU so there is no more to say. You only need the value of ΔU from process 1.