The P-V diagram of 2,g of helium gas for a ce | vedclass

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(B) The heat given to the gas is given by the first law of thermodynamics: $\Delta Q = \Delta U + \Delta W$.First,calculate the work done $\Delta W$ d

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$6$

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(B) The heat given to the gas is given by the first law of thermodynamics: $\Delta Q = \Delta U + \Delta W$.First,calculate the work done $\Delta W$ during the process $A 	o B$,which is the area under the $P-V$ curve:$\Delta W = 	ext{Area of trapezium} = \frac{1}{2} 	imes (P_A + P_B) 	imes (V_B - V_A)$$\Delta W = \frac{1}{2} 	imes (P_0 + 2P_0) 	imes (2V_0 - V_0) = \frac{1}{2} 	imes 3P_0 	imes V_0 = 1.5 P_0 V_0$.Next,calculate the change in internal energy $\Delta U = n C_v \Delta T$. For a monoatomic gas like helium,$C_v = \frac{3}{2}R$.Using the ideal gas equation $PV = nRT$,we have $nRT_A = P_0 V_0$ and $nRT_B = (2P_0)(2V_0) = 4P_0 V_0$.Thus,$nR \Delta T = nR(T_B - T_A) = 4P_0 V_0 - P_0 V_0 = 3P_0 V_0$.Therefore,$\Delta U = \frac{3}{2} (nR \Delta T) = \frac{3}{2} (3P_0 V_0) = 4.5 P_0 V_0$.Finally,the heat given is $\Delta Q = \Delta U + \Delta W = 4.5 P_0 V_0 + 1.5 P_0 V_0 = 6 P_0 V_0$.

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

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