A poly-atomic molecule (C_v = 3R, C_p = 4R,wh | vedclass

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(B) For an adiabatic process $A 	o B$,the heat exchanged $\Delta Q_{AB} = 0$.For an isothermal process $B 	o C$,the heat exchanged $\Delta Q_{BC}$ i

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$450 R (\ln 4 - \ln 3)$

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(B) For an adiabatic process $A 	o B$,the heat exchanged $\Delta Q_{AB} = 0$.For an isothermal process $B 	o C$,the heat exchanged $\Delta Q_{BC}$ is equal to the work done $W_{BC}$.Given that the process $B 	o C$ is isothermal at temperature $T = 450 \ K$ (from the graph),the heat absorbed per mole is:$\Delta Q = \Delta Q_{BC} = nRT \ln \left(\frac{V_C}{V_B}ight)$Since we are calculating heat per unit mole,$n = 1$.$\Delta Q = (1) 	imes R 	imes 450 	imes \ln \left(\frac{8 	imes 10^{-6}}{6 	imes 10^{-6}}ight)$$\Delta Q = 450 R \ln \left(\frac{4}{3}ight)$$\Delta Q = 450 R (\ln 4 - \ln 3)$.

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