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Question

One mole of a monoatomic gas is heated using PV32=const. Find the amount of heat obtained in the process, when the temperature changes by ΔT=26 K

A
92 J
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B
108 J
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C
127 J
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D
256 J
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Solution

The correct option is B 108 J
Given:
Number of moles of the gas n=1
Change in temperature ΔT=26 K
For monoatomic gas, γ=53
The process is given as PV32=constant
Comparing with PVx=constant, we get x=32
Specific heat capacity of the process C=Rγ1+R1x
C=R531+R132=R2
Amount of heat involved in the process Q=nCΔT
Q=1×(8.312)×(26) (R=8.31J K1mol1)
Q=108 J

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