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(B) The total heat supplied $(\Delta Q)$ is used for two purposes: increasing the internal energy $(\Delta U)$ and doing work against external pressur

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$500$

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(B) The total heat supplied $(\Delta Q)$ is used for two purposes: increasing the internal energy $(\Delta U)$ and doing work against external pressure $(\Delta W)$.Given:$\Delta Q = 540 \ 	ext{cal}$$P = 1.013 	imes 10^5 \ 	ext{Nm}^{-2}$$V_1 = 1 \ 	ext{cc} = 1 	imes 10^{-6} \ 	ext{m}^3$$V_2 = 1671 \ 	ext{cc} = 1671 	imes 10^{-6} \ 	ext{m}^3$$J = 4.18 \ 	ext{J/cal}$Work done against atmospheric pressure is $\Delta W = P(V_2 - V_1)$.$\Delta W = 1.013 	imes 10^5 	imes (1671 - 1) 	imes 10^{-6} \ 	ext{J}$$\Delta W = 1.013 	imes 1670 	imes 10^{-1} \ 	ext{J} \approx 169.17 \ 	ext{J}$.Converting work into calories: $\Delta W_{	ext{cal}} = \frac{169.17}{4.18} \approx 40.47 \ 	ext{cal}$.The heat required to overcome molecular attraction is the change in internal energy $\Delta U$.$\Delta U = \Delta Q - \Delta W_{	ext{cal}}$$\Delta U = 540 - 40.47 = 499.53 \ 	ext{cal} \approx 500 \ 	ext{cal}$.

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