Question

The figure shows a pV diagram for a gas going through a cycle from A to B to C and back to A. From point A to point B, the gas absorbs 50 J of heat and finds its internal (thermal) energy has increased by 20 J. Going from B to C, the internal (thermal) energy decreases by 5.0 J.

(a) How much work was done by the gas from A to B?

(b) How much work was done by the gas going from B to C?

(c) How much heat was absorbed by the gas from B to C?

Answer 1

(a)During A to B dQ=50J and dE=20J

Then by first law of thermodynamics,

dQ=dE +dW

It give us, dW=50-20=30J, therefore 30J work was done duri g the process A to B.

(b) since here in process B to C, internal energy decreases.

This decrease in internal energy will be responsible to do the work by the system. And it is equal to decrease in internal energy. Therefore, dW=5J.

(c) great will not be absorbed while there is decrease in internal energy. Therefore, dQ=0.