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Question

A certain amount of an ideal gas is initially at a pressure P1 and temperature T1. First, it undergoes an isobaric process 1 - 2 such that T2=3T14. Then, it undergoes an isochoric process 23 such that T3=T22. The ratio of the final volume to the initial volume of the ideal gas is

A
0.25
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B
1
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C
1.5
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D
0.75
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Solution

The correct option is D 0.75

Given that
P1 = Initial pressure
T1 = Initial Temperature
V3V1=??
Process 1 - 2 is constant pressure process
T2=3T14 (Given) T2T1=34
For constant pressure in ideal gas,
VT (Charle's law)
V2V1=T2T1=34(1)
Process 2 -3 is constant volume process.
T22=T3 (Given) T3T2=12
Now,
V3V1=final volumeInitial volume=V2V1
[V3=V2]
=34=0.75 [from eq (1)]

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