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Question

An ideal gas is taken through a cycle 1231 (see figure) and the efficiency of the cycle was found to be 25%. When the same gas goes through the cycle 1341, the efficiency is 10%. Find the efficiency of the cycle 12341.


A
18.6 %
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B
23.2 %
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C
32.5 %
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D
44 %
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Solution

The correct option is C 32.5 %
Cycle 1-2-3-1
From 12, work done is +ve ( volume increases) and ΔU is also positive ( temperature increases).
Similarly, for 23, W as well as ΔU are positive.
Thus, in both the processes 12 and 23, the gas absorbs heat.
Let the total heat absorbed in the process 123 be Q1. Similarly, one can argue that the gas rejects heat (say Q0) to the surrounding in the process 31. Also, let work done in the cycle be W1.


Using first law of thermodynamics for complete cycle -
ΔQ=ΔU+W [ΔUcycle=0]
Q1Q0=W1 (i)
And efficiency η=W1Q1
14=W1Q1 (ii)

Cycle 1-3-4-1


Q0= heat gained by the gas in process 13.
[due to symmetry, it will be the same as heat rejected in previous]
Let Q2= heat rejected by the gas in process 341
Let W2= work done in cycle.
η=W2Q0 110=W2Q1W1 [using (i)]
Q1W1=10W2
Q1=W1+10W2 (iii)
Using (ii) and (iii)
4W1=W1+10W2
3W1=10W2 (iv)

Efficiency of cycle 1-2-3-4-1 is given by:


η=W1+W2Q1=W1+W2W1+10W2
=1+W2W11+10W2W1=1+3101+3 [using (iv)]
=1340
i.e η=1340×100=32.5%

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