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 B108 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=Rγ−1+R1−x C=R53−1+R1−32=−R2
Amount of heat involved in the process Q=nCΔT ∴Q=1×(−8.312)×(−26)(∵R=8.31J K−1mol−1) ⇒Q=108 J