Question

An ideal gas goes from state a to state b through
two process I and II as shown.
The heat absorbed and work done in process I are
100 J and 80 J respectively. If work done in
process Il is 60 J, then heat absorbed in process
Il will be
(1) 90 J
(2) 85 J
(3) 80 J
(4) 75 J​

Answer 1

Answer:

(3) 80 J

Explanation:

If we assume that the following diagram .

For process I

Q= 100 J ,W= 80 J

For process II

W= 60 J ,Q= ?

As we know that internal energy of a system is a point function .It means that it only depends only on the initial and final position.

From first law of thermodynamics

Q = ΔU +W

So

100 = ΔU + 80                --------------1

Q = ΔU + 60                 -----------2

From 1 and 2

100  - Q = 80 - 60

100 - Q = 20

Q = 100 -20 = 80 J

Q = 80 J

(3) 80 J