Heat exchanged in a cyclic process

Classical Mechanics Level 3

One mole of ideal monoatomic gas is taken round the cyclic process A B C A ABCA . The following is a schematic P V PV graph of a process A B C A ABCA .

Find the heat exchange in the process A B AB .

9 P 0 V 0 2 \dfrac{9P_0V_0}{2} 3 P 0 V 0 3P_0V_0 − 3 P 0 V 0 -3P_0V_0 3 P 0 V 0 2 \dfrac{3P_0V_0}{2}

Md Zuhair by Md Zuhair , 20, India

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1 solution

Ayon Ghosh
Apr 22, 2017

In an isochoric process,

d Q dQ = d U dU + + d W dW

( here d is short for delta not differential)

But d U dU = u C v d T u C_v dT = C v d T C_v dT

(u is short for mu - no. of moles =1 )

Also d W dW = P d V PdV = 0 0 since d V dV = 0 0 for an isochoric process.

Hence d Q dQ = d U dU = C v d T C_v dT

From Ideal gas equation, we get,

d T dT = d P V dPV / / R R ( since u =1 )

Substituting,

d Q dQ = C v d P V C_v dPV / / R R = 3 / 2 R ∗ 2 P ∗ V 3/2 R * 2P * V / / R R = 3 P V 3PV

( because for monoatomic ideal gases gamma = C p C_p / / C v C_v = 5 / 3 5/3 and also C p C_p - C v C_v = R R from which we get C p C_p = 3 / 2 3/2 )

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