An ideal monatomic gas of n moles is taken th | vedclass

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(C) For a monatomic gas,the adiabatic index $\gamma = 5/3$. In the $V-T$ diagram,the process $XY$ is a straight line passing through the origin,which

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$(1.60)$

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(C) For a monatomic gas,the adiabatic index $\gamma = 5/3$. In the $V-T$ diagram,the process $XY$ is a straight line passing through the origin,which represents an isobaric process $(V \propto T)$. The process $YZ$ is adiabatic $(TV^{\gamma-1} = 	ext{constant})$,and $ZW$ is isobaric. Given $V_W = 64 \ cm^3, V_X = 125 \ cm^3, V_Y = 250 \ cm^3$. Since $XY$ is isobaric,$P_X = P_Y$. For the adiabatic process $YZ$,$T_Y V_Y^{\gamma-1} = T_Z V_Z^{\gamma-1}$. For the adiabatic process $WX$,$T_W V_W^{\gamma-1} = T_X V_X^{\gamma-1}$. Since $XY$ is isobaric,$V_X/T_X = V_Y/T_Y \Rightarrow T_Y = T_X (V_Y/V_X) = T_X (250/125) = 2T_X$. Heat absorbed in isobaric process $XY$ is $Q = nC_P \Delta T = n(5R/2)(T_Y - T_X) = (5/2)nR(2T_X - T_X) = (5/2)nRT_X$. From the adiabatic process $WX$,$T_X = T_W (V_W/V_X)^{\gamma-1} = T_W (64/125)^{5/3-1} = T_W (64/125)^{2/3} = T_W (4/5)^2 = T_W (16/25)$. Thus,$Q = (5/2) nRT_W (16/25) = (5/2) (1 \ J) (16/25) = (1/2) (16/5) \ J = 8/5 \ J = 1.6 \ J$.

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