Question

Initially at a temperature of 90.0 ∘C, 0.220 m3 of air expands at a constant gauge pressure of 1.31×105 Pa to a volume of 1.50 m3 and then expands further adiabatically to a final volume of 2.35 m3 and a final gauge pressure of 2.23×104 Pa. Compute the total work done by the air. CV for air is 20.8 J/(mol⋅K) .

Answer 1

initial temperature=T1=273+90=

363 K

initial volume=V1=0.22 m^3

initial pressure P1=1.31*10^5+101325=232325 Pa

at the end of first process,

pressure=P2=P1=232325 Pa

volume=V2=1.5 m^2

then work done=P1*(V2-V1)=232325*(1.5-0.22)=297376 J

second process:

at the beginning of the second process which is adiabatic in nature,

initial volume=V2=1.5 m^3

initial pressure=P2=232325 Pa

final volume=V3=2.35 m^3

final pressure=P3=2.23*10^4+101325=123625 Pa

work done in adabatic process=(P3*V3-P2*V2)/(1-gamma)

where gamma=Cp/Cv

=(Cv+R)/Cv

=(20.8+8.314)/20.8=1.3997

hence work done=(123625*2.35-232325*1.5)/(1-1.3997)

=145030.648 J

hence total work done=297376 +145030.648=442406.6480 J