Solution 36
We analyze the two processes separately for 1 mole of monoatomic gas ($n=1, C_v = \frac{3}{2}R$).
Process A $\to$ B: Given $P \propto T$.
From the ideal gas equation $PV = RT$, if $P = kT$, then $kT \cdot V = RT \implies V = \frac{R}{k} = \text{constant}$. This is an isochoric process ($W_{AB} = 0$).
$$ \Delta Q_{AB} = n C_v (T_B – T_A) = \frac{3}{2}R(T_B – T_A) $$Process B $\to$ C: Given $P \propto \sqrt{T}$, so $P = c T^{1/2}$.
Substitute $P = RT/V$:
$$ \frac{RT}{V} = c T^{1/2} \implies V = \frac{R}{c} T^{1/2} $$To find the work done, we differentiate: $dV = \frac{R}{2c} T^{-1/2} dT$.
$$ dW = P dV = (c T^{1/2}) \left( \frac{R}{2c} T^{-1/2} dT \right) = \frac{R}{2} dT $$Integrating from $T_B$ to $T_C$:
$$ W_{BC} = \frac{R}{2}(T_C – T_B) $$The change in internal energy is $\Delta U_{BC} = \frac{3}{2}R(T_C – T_B)$. Thus, heat supplied is:
$$ \Delta Q_{BC} = W_{BC} + \Delta U_{BC} = \frac{R}{2}(T_C – T_B) + \frac{3}{2}R(T_C – T_B) = 2R(T_C – T_B) $$Total Heat Supplied:
$$ \Delta Q_{total} = \frac{3}{2}R(T_B – T_A) + 2R(T_C – T_B) $$ $$ = \frac{3}{2}RT_B – \frac{3}{2}RT_A + 2RT_C – 2RT_B $$ $$ = 2RT_C – \frac{1}{2}RT_B – \frac{3}{2}RT_A $$
$$ \Delta Q_{total} = \frac{R(4T_C – T_B – 3T_A)}{2} $$