The P-V diagram shown below indicates two pat | vedclass

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(A) According to the first law of thermodynamics, $Q = \Delta U + W$, where $Q$ is the heat transferred to the system, $\Delta U$ is the change in int

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$11PV/2$

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(A) According to the first law of thermodynamics, $Q = \Delta U + W$, where $Q$ is the heat transferred to the system, $\Delta U$ is the change in internal energy, and $W$ is the work done by the system.For any process between two states, $\Delta U$ is a state function and remains the same for both paths.For Path-$1$ (isobaric process at $P$ from $V$ to $3V$):$W_1 = P(3V - V) = 2PV$.Given $Q_1 = 5PV$, then $\Delta U = Q_1 - W_1 = 5PV - 2PV = 3PV$.For Path-$2$ (consisting of an isochoric process and an isobaric process):Work done $W_2$ is the area under the path in the $P-V$ diagram.$W_2 = 	ext{Area of rectangle} + 	ext{Area of triangle} = (2V)(P) + \frac{1}{2}(2V)(\frac{3}{2}P - P) = 2PV + \frac{1}{2}(2V)(\frac{1}{2}P) = 2PV + 0.5PV = 2.5PV$.Since $\Delta U$ is the same for both paths, $\Delta U = 3PV$.Thus, $Q_2 = \Delta U + W_2 = 3PV + 2.5PV = 5.5PV = 11PV/2$.

The $P-V$ diagram shown below indicates two paths along which a sample of gas can be taken from state $A$ to state $B$. The energy equal to $5PV$ in the form of heat is required to be transferred if Path-$1$ is chosen. How much energy in the form of heat should be transferred if Path-$2$ is chosen?

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