Question

150 J of heat is added to a gaseous system whose internal energy is 10 cal, then the amount of external work done is:

(A) 42 J
(B) 192 J
(C) 108 J
(D) 10 J ​

Answer 1

Answer:

The first law of thermodynamics states that-

dq= du + dw

( where dq= heat added, -dq= heat taken away, dw= work done by the system, -dw= work done on the system )

Now, heat is added. We need to find work. Using the first law,

150= (10 × 4.2) + dw ( as 1 cal = 4.2 J)

⇒ dw= 150 - 42

⇒ dw= 108 J