The P-V diagram of 2,g of He gas for the proc | vedclass

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(B) From the ideal gas equation $PV = \mu RT$,we have:At point $A$: $P_A V_A = P_0 V_0 = \mu R T_A$At point $B$: $P_B V_B = (2P_0)(2V_0) = 4P_0 V_0 =

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$6$

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(B) From the ideal gas equation $PV = \mu RT$,we have:At point $A$: $P_A V_A = P_0 V_0 = \mu R T_A$At point $B$: $P_B V_B = (2P_0)(2V_0) = 4P_0 V_0 = \mu R T_B$Change in internal energy $\Delta U = \mu C_v \Delta T = \frac{\mu R \Delta T}{\gamma - 1}$. For monoatomic $He$ gas,$C_v = \frac{3}{2}R$,so $\Delta U = \frac{3}{2} \mu R (T_B - T_A) = \frac{3}{2} (P_B V_B - P_A V_A) = \frac{3}{2} (4P_0 V_0 - P_0 V_0) = \frac{3}{2} (3P_0 V_0) = 4.5 P_0 V_0$.Work done $\Delta W$ is the area under the $P-V$ graph,which is a trapezoid: $\Delta W = \frac{1}{2} (P_A + P_B) (V_B - V_A) = \frac{1}{2} (P_0 + 2P_0) (2V_0 - V_0) = \frac{1}{2} (3P_0)(V_0) = 1.5 P_0 V_0$.According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W = 4.5 P_0 V_0 + 1.5 P_0 V_0 = 6 P_0 V_0$.

The $P-V$ diagram of $2\,g$ of $He$ gas for the process $A \to B$ is shown. What is the heat given to the gas (in $,P_0V_0$)?

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