Question

One mole of an ideal monoatomic gas is taken through a thermodynamic process shown in the P – V diagram. The heat supplied to the system in this process is K × ($\pi$ + 10)$P_0 V_0$. Determine the value of K.

Answer 1

1

Answer 2

The process is a cycle A $\to$ B $\to$ A.

Process A to B is isobaric expansion at $P=2P_0$ from $V=2V_0$ to $V=4V_0$. Work done $W_{AB} = 2P_0 (4V_0 - 2V_0) = 4P_0 V_0$. Change in internal energy $\Delta U_{AB} = n C_v \Delta T_{AB}$. $T_A = \frac{(2P_0)(2V_0)}{R} = \frac{4P_0 V_0}{R}$, $T_B = \frac{(2P_0)(4V_0)}{R} = \frac{8P_0 V_0}{R}$. $\Delta T_{AB} = \frac{4P_0 V_0}{R}$. For monoatomic gas, $C_v = \frac{3}{2}R$. $\Delta U_{AB} = 1 \times \frac{3}{2}R \times \frac{4P_0 V_0}{R} = 6 P_0 V_0$. Heat supplied $Q_{AB} = \Delta U_{AB} + W_{AB} = 6 P_0 V_0 + 4P_0 V_0 = 10 P_0 V_0$.

Process B to A is a curved path from $(4V_0, 2P_0)$ to $(2V_0, 2P_0)$ passing through $(3V_0, 4P_0)$. This is the upper half of an ellipse centered at $(3V_0, 2P_0)$ with semi-axes $a=V_0$ and $b=2P_0$. Work done $W_{BA} = \int_{4V_0}^{2V_0} P dV$. The area under the curve is the area of the rectangle with height $2P_0$ and width $2V_0$ plus the area of the semi-ellipse with semi-axes $a=V_0$ and $b=2P_0$. The area of the semi-ellipse is $\frac{1}{2} \pi a b = \frac{1}{2} \pi (V_0)(2P_0) = \pi P_0 V_0$. The area under the curve from $V=4V_0$ to $V=2V_0$ is the area of the rectangle from $V=2V_0$ to $V=4V_0$ at $P=2P_0$ plus the area of the semi-ellipse above $P=2P_0$. The area under the curve is $2P_0 (4V_0 - 2V_0) + \pi P_0 V_0 = 4P_0 V_0 + \pi P_0 V_0$. Since the volume is decreasing, the work done by the system is negative. $W_{BA} = -(4P_0 V_0 + \pi P_0 V_0) = -4P_0 V_0 - \pi P_0 V_0$. Change in internal energy $\Delta U_{BA} = n C_v \Delta T_{BA}$. $\Delta T_{BA} = T_A - T_B = - \frac{4P_0 V_0}{R}$. $\Delta U_{BA} = 1 \times \frac{3}{2}R \times (-\frac{4P_0 V_0}{R}) = -6 P_0 V_0$. Heat supplied $Q_{BA} = \Delta U_{BA} + W_{BA} = -6 P_0 V_0 + (-4P_0 V_0 - \pi P_0 V_0) = -10 P_0 V_0 - \pi P_0 V_0$.

The heat supplied to the system in this process (the cycle) is the net heat transfer $Q_{cycle} = Q_{AB} + Q_{BA} = 10 P_0 V_0 + (-10 P_0 V_0 - \pi P_0 V_0) = -\pi P_0 V_0$. The given heat supplied is $K \times (\pi + 10)P_0 V_0$. If this is the net heat supplied, then $-\pi P_0 V_0 = K (\pi + 10) P_0 V_0$, so $K = -\frac{\pi}{\pi + 10}$.

If the question asks for the magnitude of the heat transferred during the process from B to A, then $|Q_{BA}| = |-10 P_0 V_0 - \pi P_0 V_0| = (10 + \pi) P_0 V_0$. If $|Q_{BA}| = K (\pi + 10) P_0 V_0$, then $(10 + \pi) P_0 V_0 = K (\pi + 10) P_0 V_0$, so $K=1$.

Given the structure of the expression, it is highly probable that the question is asking for the magnitude of the heat transferred during the process from B to A.

The final answer is $\boxed{1}$.