Heat - 1

Glad to see problems on thermodynamics. Thank you for uploading. The following solution focusses more on the method rather than the concepts themselves.

For a monoatomic gas, $C_v = \frac{3R}{2}$

Applying the first law of thermodynamics (FLT) for the process $AB$ gives:

$\Delta Q_{AB} = \Delta U_{AB} + \Delta W_{AB}$

Since $AB$ is an isochoric process, the work done by the gas $\Delta W_{AB}=0$

$\Delta Q_{AB} = \Delta U_{AB} = n C_v (T_B - T_A) \dots (1)$

Now, for process $BC$ , which is isobaric, applying FLT gives:

$\Delta Q_{BC} = \Delta U_{BC} + \Delta W_{BC} \implies = n C_v (T_C - T_B) + 2P(2V - V) \dots (2)$

Now applying the ideal gas law at points $A$ , $B$ and $C$ gives:

$P_AV_A = nRT_A \implies PV = nRT_A$ $P_BV_B = nRT_B \implies 2PV = nRT_B$ $P_CV_C = nRT_C \implies 4PV = nRT_C$

Using these relations and substituting into (1) and (2), adding them and simplifying gives:

$\boxed{\Delta Q_{AB} + \Delta Q_{BC}= \frac{13PV}{2}}$ Now, the question demands the head added by the source into the cycle. In processes $AB$ and $BC$ , these values are positive. However, doing similar calculations for processes $CD$ and $DA$ will show that heat is rejected by the closed system. Hence the heat exchanges in these processes are not to be accounted for.