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Q.

One mole of an ideal gas undergoes a process.P=P01+V0V2Here P0 and V0 are constants. Change in temperature of the gas when volume is changed from V=V0 to V=2V0is

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a

P0V0

b

2P0V05R

c

11P0V010R

d

5P0V04R

answer is B.

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Detailed Solution

At V=V0:P=P02
Ti=PVnR=P02V0R=P0V02R(n=1) and at V=2V0:P=4P05 Tf=PVnR=2V04P05R=8P0V05R ΔT=TfTi=8512P0V0R

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