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- Homework Statement
- In the thermodynamic cycle shown, one mole of monoatomic ideal gas is taken through the cycle. AB is a reversible isothermal expansion at a temperature of 800K in which the volume of the gas is doubled. BC is an isobaric contraction to the original volume in which the temperature is reduced to 300K. CA is a constant volume process in which the pressure and temperature return to their initial values. Find the net amount of heat absorbed .

- Relevant Equations
- PV=RT

Q=C(T2-T1)

dU=dQ-dW

Cv=1.5R for monoatomic ideal gas

Cp = Cv+R = 2.5R

In AB, dU=0 (Isothermal)

Hence dQ=dW

W1= RT ln(2V/V) = RT ln(2) = 800R*ln(2)

Q1=W1 = 800R*ln(2)

Q2= Cp(300-800)=-500Cp = -500*2.5R

Q3 = Cv(800-300) = 500Cv = 500*1.5R

So Q= 800R*ln(2) -500*2.5R + 500*1.5R = 800R*ln(2) -500R

**Hence net heat absorbed is 800R*ln(2) -500R**

But.

Suppose I find work done instead:

W1 = 800R*ln(2)

W2 = P2(ΔV) = P2*(V-2V)=-P2*V

W3=0

Now, P1V1=RT

P2V2 = RT

So, P2=P1/2=0.5P1

So,W2=-0.5*P1*V1=-0.5*R*T1 = -400R

So, Net work done is 800*R*ln(2) - 400R

As this is a cycle, dU=0 over it. And net heat absorbed is net work done. Hence

**Net heat absorbed is 800*R*ln(2) - 400R**

There is inconsistency here. What did I do wrong?