Question

When a system is taken from state i to state f along the path iaf , it is found that Q = 50cal and W = 20cal. along the path ibf , Q = 36cal W along the path ibf is [AIEEE 2007] ​

Answer 1

Explanation:- From first law of thermodynamics ,

∆W = ∆U + W

For path iaf , given Q = 50 and W = 20

• 50 = ∆U + 20
• ∆U = 50- 20
• ∆U = 30

Now , for path ibf , given Q = 36 and we calculated ∆U = 30cal .

∆Q = ∆U + W

• W =Q - ∆U
• W = 36- 30 = 6 cal

Since , W along the path ibf is 6cal

Answer 2

Answer:

By first law of thermodynamics :

Q=ΔU+W

In path iaf:

50=ΔU+20

∴ΔU=30

As this change in internal energy is between i & f and does not depend on path.

Therefore along path ibf:

Q=ΔU+W

36=30+W

∴W=6 cal

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