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Question

# 1 mole of an ideal gas, initially at 400 K and 10 atm is first expanded at constant pressure till the volume is doubled. Then the gas is made to undergo an isochoric process, in which its temperature is found to decrease. In the last step, the gas was compressed reversibly and adiabatically to its initial state. Determine the net work involved in this cyclic process (in terms of R). Given, CV for gas =1.5 R, (4)−1/3=0.63. If |W|=2 R×z, what is the value of z?

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Solution

## (1) (10 atm, 400 K, 1 mole) vdoubles→ v constant but T→(3)Adiabatic−−−−−−−−−−−−−−→reversible compression(4)→sameas(1) Now, V(1)=nRT(1)P(1)=1×R×40010=40R V(2)=2×V(1)=80R=V(3) P - V curve is : now, (3) → (1) : Adiabatic reversible. So, P(1)V(1)=P(3)V(3)γ (here,γ=53) ∴ 10×(40 R)83=P(3)(80 R)53 ⇒ P(3)=10×(40R80R)53=10253=5413 or, P(3)=5×0.63=3.15atm. ωtotal=ω(1)→(2)+ω(2)→(3)+ω(3)→(1) ω(1)→(2)=10 atm(80R−40R)=−400R ω(2)→(3)=zero(isochoric) ​​​​​ω(3)→(1)=△E(adiabatic) ​=P(1)V(1)−P(3)V(3)γ−1=32[400R−252R]​​ =222R.​​​​​​ ∴ ωtotal=−400R+222R=−178R ∴|ω|=178R=2R×89=2R×Z ∴Z=89

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