An open flask at 27^{circ}C contains air. To | vedclass

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(C) For an open flask,pressure $(P)$ and volume $(V)$ remain constant.According to the ideal gas law,$PV = nRT$,which implies $n \propto \frac{1}{T}$

Vedclass · Accepted Answer

$177$

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(C) For an open flask,pressure $(P)$ and volume $(V)$ remain constant.According to the ideal gas law,$PV = nRT$,which implies $n \propto \frac{1}{T}$ at constant $P$ and $V$.Given: $T_1 = 27^{\circ}C = 300 \, K$.Let the initial number of moles be $n_1 = n$.After heating,one-third of the air is expelled,so the remaining moles are $n_2 = n - \frac{n}{3} = \frac{2n}{3}$.Using the relation $\frac{n_1}{n_2} = \frac{T_2}{T_1}$:$\frac{n}{2n/3} = \frac{T_2}{300}$$\frac{3}{2} = \frac{T_2}{300}$$T_2 = \frac{3}{2} 	imes 300 = 450 \, K$.Converting to Celsius: $T_2 = 450 - 273 = 177^{\circ}C$.

An open flask at $27^{\circ}C$ contains air. To what temperature $(^{\circ}C)$ should it be heated so that one-third of the air is expelled from the flask?

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