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Question

# One mole of an ideal monoatomic gas undergoes the process P=αT1/2, where α is a constant. Find the work done by the gas if its temperature increases by 50K.

A
407.75J
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B
207.75J
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C
307.75J
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D
107.75J
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Solution

## The correct option is C 207.75JThe given process is P=αT1/2⟹ PT−1/2=α=constantComparing with PTm1−m=constant we get m1−m=−12⟹ m=−1For monoatomic gas γ=1.67Molar specific heat of the gas C=Rγ−1+R1−m∴ C=R1.67−1+R1−(−1)=2RMolar specific heat of the gas C=2R Given : ΔT=50K Heat supplied to the gas Q=nCΔT where n=1∴ Q=2R×50=100RChange in internal energy ΔU=nRΔTγ−1∴ ΔU=1R×501.67−1=75RFrom 1st law Q=ΔU+W∴ 100R=75R+W⟹W=25R=25×8.314=207.75 J

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