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

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(B) The work done $W$ is equal to the area under the $P-V$ graph.$W = 	ext{Area of trapezium} = \frac{1}{2} 	imes (P_A + P_B) 	imes (V_B - V_A)$$W

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$6$

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(B) The work done $W$ is equal to the area under the $P-V$ graph.$W = 	ext{Area of trapezium} = \frac{1}{2} 	imes (P_A + P_B) 	imes (V_B - V_A)$$W = \frac{1}{2} 	imes (P_0 + 2P_0) 	imes (2V_0 - V_0) = \frac{1}{2} 	imes (3P_0) 	imes (V_0) = \frac{3}{2} P_0V_0$Using the ideal gas equation $PV = \mu RT$:At point $A$: $P_0V_0 = \mu RT_A$At point $B$: $(2P_0)(2V_0) = 4P_0V_0 = \mu RT_B$Change in internal energy $\Delta U = \mu C_v \Delta T = \mu \left( \frac{3}{2}R ight) (T_B - T_A) = \frac{3}{2} (\mu RT_B - \mu RT_A)$Substituting the values: $\Delta U = \frac{3}{2} (4P_0V_0 - P_0V_0) = \frac{3}{2} (3P_0V_0) = \frac{9}{2} P_0V_0$According to the first law of thermodynamics,$Q = W + \Delta U$$Q = \frac{3}{2} P_0V_0 + \frac{9}{2} P_0V_0 = \frac{12}{2} P_0V_0 = 6 P_0V_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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