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Question

# Two moles of an ideal monoatomic gas occupy a volume 2V at temperature 300 K, it expands to a volume 4V adiabatically, then the final temperature of gas is

A
179 K
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B
189 K
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C
199 K
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D
219 K
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Solution

## The correct option is B 189 KGiven,n=2V1=2VV2=4VT1=300KFor ideal monoatomic gas, γ=1.66In adiabatic process, PVγ=k(constant). . . . . .(1)From ideal gas equation,PV=nRT=2RTP=2RTVFrom equation (1), 2RTVVγ=kTVγ−1=k′(constant)T1Vγ−11=T2Vγ−12T2=T1(V1V2)γ−1T2=300×(2V4V)1.66−1T2=300×0.630T2=189KThe final temperature of gas is 189K.The correct option is B.

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