1
Question
A gas is enclosed in a cylinder with a movable frictionless piston. Its initial thermodynamic state at pressure Pi=105 Pa and volume Vi=10−3 Pa changes to a final state at Pf=(132)×105 Pa and Vf=8×10−3 m3 in an adiabatic quasi-static process, such that P3V5= constant. Consider another thermodynamic process that brings the system from the same initial state to the same final state in two steps : an isobaric expansion at Pi followed by an isochoric (isovolumetric) process at volume Vf. The amount of heat supplied to the system in the two-step process is approximately
A gas is enclosed in a cylinder with a movable frictionless piston. Its initial thermodynamic state at pressure Pi=105 Pa and volume Vi=10−3 Pa changes to a final state at Pf=(132)×105 Pa and Vf=8×10−3 m3 in an adiabatic quasi-static process, such that P3V5= constant. Consider another thermodynamic process that brings the system from the same initial state to the same final state in two steps : an isobaric expansion at Pi followed by an isochoric (isovolumetric) process at volume Vf. The amount of heat supplied to the system in the two-step process is approximately
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Solution
The correct option is C 588 J
In adiabatic process ideal gas equation: PVγ=Constant
from given data, P3V5=constant
So, PV53=k
where, γ=53
Total work done during process 1 and 2,
Wa=PfVf−PiVi1−γ→Wa=132×10−5×8×10−3−105×10−31−5/3Wa=112.5 J
In adiabatic process the internal energy of system ΔQ=0so,ΔU=−W⇒−112.5 J
Total heat supply between 1 and 2
q1+q2=ΔU+Pi(Vf−Vi)=−112.5+105(8−1)×10−3=587.5 J
In adiabatic process ideal gas equation: PVγ=Constant
from given data, P3V5=constant
So, PV53=k
where, γ=53
Total work done during process 1 and 2,
Wa=PfVf−PiVi1−γ→Wa=132×10−5×8×10−3−105×10−31−5/3Wa=112.5 J
In adiabatic process the internal energy of system ΔQ=0so,ΔU=−W⇒−112.5 J
Total heat supply between 1 and 2
q1+q2=ΔU+Pi(Vf−Vi)=−112.5+105(8−1)×10−3=587.5 J
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