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Question

# In a thermodynamic process helium gas obeys the law TP−25=constant. If temperature of 2 moles of the gas is raised from T to 3T, then

A

Heat given to the gas is 9RT

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B

Heat given to the gas is zero

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C

Increase in internal energy is 6RT

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D

Work done by the gas is - 6RT

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Solution

## The correct options are B Heat given to the gas is zero C Increase in internal energy is 6RT D Work done by the gas is - 6RT TP−25=constant ∴(PV)P−25=constant or PV53=constant Comparing with the equation, PVx=constant We have x=53 Molar heat capacity of equation (1) is given by C=Cv+R1−x =32R+R1−53=0 ∴Q=nCΔT=0 ΔU=−W =nCvΔT =(2)(32R)(3T−T) =6RT

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