A diatomic gas goes through a process ΔU+λW=0, now for different value of λ which of the following is incorrect.(W = work done by the gas)
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a
λ=1,then process is adiabatic
b
λ=0,then process is isochoric
c
λ=−52,then the process is isobaric
d
λ=52,then the process follows PV2= constant
answer is B.
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Detailed Solution
Forλ=1,dU+W=0⇒Q=0 For λ=0,dU=0⇒T= cons tan t isothermal, incorrect choice (2)for λ=−52, dU=52WQ=dU+W=72WQdU=75⇒Q=75×n52RdTQ=nCpdT⇒pressure is constFor λ=52⇒W=−25dU=−25nCvdTW=−nRdT=P2V2−P1V1−1 if PVN= cons tant⇒W=∫i1n PdV=P2V2−P1V11−NSolving 1−N=−1⇒N=2