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On the P-V diagram, two processes ‘acb’ and ‘adb’ are drawn as shown in the given diagram. For path acb, if Q (heat given to the gas) = 200 J, then W (work done) = 80 J. For path 'adb', if Q = 144 J, then find work done.

On the P-V diagram, two processes ‘acb’ and ‘adb’ are drawn as shown in the given diagram. For path acb, if Q (heat given to the gas) = 200 J, then W (work done) = 80 J. For path 'adb', if Q = 144 J, then find work done. Correct Answer 24 J  

Concept:

The First Law of Thermodynamics for a process (state 1 to state 2) is given by-

Q1-2 = ΔU + W1-2

Q1-2 = (U2 - U1) + W1-2

Calculation:

Given:

Process acb:

Qa-c-b = 200 J, Wa-c-b = 80 J

Process adb:

Qa-d-b = 144 J, Wa-d-b = ?

Process acb:

Qa-c-b = Ub - Ua + Wa-c-b

∴ Ub - Ua = Qa-c-b - Wa-c-b

Process adb:

Qa-d-b = Ub - Ua + Wa-d-b

∴ Ub - Ua = Qa-d-b - Wa-d-b

∴ Qa-c-b - Wa-c-b = Qa-d-b - Wa-d-b

∴ 200 - 80 = 144 - Wa-d-b

∴ Wa-d-b = 24 J.

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