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

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(B) For a monoatomic gas like helium,the degree of freedom $f = 3$.The change in internal energy $\Delta U$ is given by:$\Delta U = \frac{f}{2}(P_f V_

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(B) For a monoatomic gas like helium,the degree of freedom $f = 3$.The change in internal energy $\Delta U$ is given by:$\Delta U = \frac{f}{2}(P_f V_f - P_i V_i) = \frac{3}{2}(2P_0 	imes 2V_0 - P_0 	imes V_0) = \frac{3}{2}(4P_0V_0 - P_0V_0) = \frac{3}{2}(3P_0V_0) = 4.5P_0V_0$.The work done $W$ in the process $A 	o B$ is equal to the area under the $P-V$ graph,which is a trapezoid:$W = 	ext{Area} = \frac{1}{2} 	imes (P_A + P_B) 	imes (V_B - V_A) = \frac{1}{2} 	imes (P_0 + 2P_0) 	imes (2V_0 - V_0) = \frac{1}{2} 	imes 3P_0 	imes V_0 = 1.5P_0V_0$.According to the first law of thermodynamics,the heat given $\Delta Q$ is:$\Delta Q = \Delta U + W = 4.5P_0V_0 + 1.5P_0V_0 = 6P_0V_0$.

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 ${P_0}{V_0}$)?

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