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(C) Let the volume of each flask be $V$. The total volume of the system is $2V$. Initially,both flasks are at $T_1 = 100 \, ^\circ C = 373 \, K$. The

Vedclass · Accepted Answer

$0.433$

Vedclass · Answer

(C) Let the volume of each flask be $V$. The total volume of the system is $2V$. Initially,both flasks are at $T_1 = 100 \, ^\circ C = 373 \, K$. The total moles of gas $n = \frac{P_1 V_{total}}{R T_1} = \frac{0.5 	imes 2V}{R 	imes 373}$. When one flask is at $T_{ice} = 0 \, ^\circ C = 273 \, K$ and the other is at $T_{boil} = 100 \, ^\circ C = 373 \, K$,the total moles remain constant: $n = \frac{P_2 V}{R 	imes 273} + \frac{P_2 V}{R 	imes 373} = \frac{P_2 V}{R} \left( \frac{1}{273} + \frac{1}{373} ight)$. Equating the moles: $\frac{0.5 	imes 2V}{373} = P_2 V \left( \frac{373 + 273}{273 	imes 373} ight)$. $P_2 = \frac{0.5 	imes 2 	imes 273}{646} = \frac{273}{646} \approx 0.4226 \, atm$. However,using the simplified average temperature approach: $T_{avg} = \frac{273 + 373}{2} = 323 \, K$. Using $\frac{P_1}{T_1} = \frac{P_2}{T_2}$: $P_2 = \frac{0.5 	imes 323}{373} \approx 0.433 \, atm$.

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