Two flasks I and II shown below are connected | vedclass

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(A) Initial moles in flask $I$: $n_I = \frac{2.8 \, g}{28 \, g \, mol^{-1}} = 0.1 \, mol$. Initial moles in flask $II$: $n_{II} = \frac{0.2 \, g}{28 \

Vedclass · Accepted Answer

$84$

Vedclass · Answer

(A) Initial moles in flask $I$: $n_I = \frac{2.8 \, g}{28 \, g \, mol^{-1}} = 0.1 \, mol$. Initial moles in flask $II$: $n_{II} = \frac{0.2 \, g}{28 \, g \, mol^{-1}} = \frac{1}{140} \, mol \approx 0.00714 \, mol$. Total moles $n_{total} = 0.1 + \frac{1}{140} = \frac{15}{140} = \frac{3}{28} \, mol$. Since the flasks are connected and the system reaches thermal equilibrium,the final temperature $T$ is determined by the heat balance: $n_I C_v (T - 300) + n_{II} C_v (T - 60) = 0$. $0.1(T - 300) + \frac{1}{140}(T - 60) = 0$. $14(T - 300) + (T - 60) = 0$ $\Rightarrow 15T = 4260$ $\Rightarrow T = 284 \, K$. Using $PV = nRT$ for the total system: $P(V_I + V_{II}) = n_{total} RT$. $P(1 \, L + 2 \, L) = (\frac{3}{28} \, mol) 	imes (0.0831 \, L \, bar \, K^{-1} \, mol^{-1}) 	imes 284 \, K$. $3P = \frac{3 	imes 0.0831 	imes 284}{28} \approx 2.528 \, bar \cdot L$. $P = 0.8427 \, bar = 84.27 	imes 10^{-2} \, bar$. Rounding to the nearest integer,$x = 84$.

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