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Question

Four identical samples of an ideal gas are taken along the paths A,B,C and D from state 1 to state 2 as shown in the PV indicator diagram. Let Q,W and ΔU represent the heat supplied, work done and change in internal energy of the gas respectively. Choose the correct option among the following :


A
QAQD=WAWD
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B
QBWB>QCWC
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C
WA<WB<WC<WD
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D
QA<QB<QC<QD
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Solution

The correct option is A QAQD=WAWD
We know that area under PV curve gives the work done during the process.
Work done by the gas in all four processes is positive and is in the order
WA>WB>WC>WD .....(1)
Hence, option (c) is wrong.
The change in internal energy ΔU is the same for all processes as initial and final states for all processes are the same.
Using first law of thermodynamics for each process A, B, C and D, we get
QA=ΔU+WA ......(2)
QB=ΔU+WB ......(3)
QC=ΔU+WC ......(4)
QD=ΔU+WD ......(5)
Hence, from (1) we can say that, QA>QB>QC>QD
So, option (d) is also wrong.

From equation (3) and (4),
QBQC=WBWC
QBWB=QCWC
i.e option (b) is wrong.

From equation (2) and (5),
QAQD=WAWD
Thus, option (a) is the correct answer.

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