When 1 mol of ice melts at 0 ^oC and a cons | vedclass

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Question

(A) Given: $q = +1440 \ cal$ (heat absorbed by the system).Pressure $P = 1 \ atm$.Change in volume $\Delta V = V_{water} - V_{ice} = 0.0180 \ L - 0.01

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$1440$

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(A) Given: $q = +1440 \ cal$ (heat absorbed by the system).Pressure $P = 1 \ atm$.Change in volume $\Delta V = V_{water} - V_{ice} = 0.0180 \ L - 0.0196 \ L = -0.0016 \ L$.Work done $w = -P \Delta V = -1 \ atm 	imes (-0.0016 \ L) = +0.0016 \ L \cdot atm$.Convert $w$ to calories: $1 \ L \cdot atm = 24.22 \ cal$.$w = 0.0016 	imes 24.22 \ cal \approx 0.0387 \ cal$.Using the first law of thermodynamics: $\Delta U = q + w$.$\Delta U = 1440 \ cal + 0.0387 \ cal \approx 1440 \ cal$ (since the work done is negligible compared to the heat absorbed).

When $1 \ mol$ of ice melts at $0 \ ^oC$ and a constant pressure of $1 \ atm$,the system absorbs $1440 \ cal$ of heat. If the molar volumes of ice and water are $0.0196 \ L$ and $0.0180 \ L$ respectively,calculate the change in internal energy $(\Delta U)$ of the system in $cal$.

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