Question

An open container of volume 3 litre contains air at 1 atmospheric pressure. The container is heated from initial temperature ${27}^{\circ }\text{C}$ to $t^{\circ }\text{C}$, the amount of the gas expelled from the container measured 1.45 litre at ${17}^{\circ }\text{C}$ and 1 atm. Find temperature $t$

Answer 1

According to Charle's law,
$V\propto T$ at constant pressure
Hence increase in temperature, increases the volume. Hence to maintain constant pressure, some volume of the gas is expelled.

Initial volume of the gas $V_1=3L$
Initial temperature of the gas $T_1={27}^{\circ }\text{C}=300K$
Final volume $V_2=\text{initial volume}+\text{volume expelled}$
$=3+\Delta VL$
Final temperature $T_2=?$
Determination of $\Delta V$ at $T_2$:
Given $\Delta VL$ at ${17}^{\circ }\text{C}$ or 290 K = 1.45 L

Accoridng to Charle's law At constant pressure $V_1{T}_2=V_2{T}_1$
$\Delta VL$ at temperature $T_2$
$\Delta V$ = $\frac{1.45\times T_2}{290}$

Hence final volume $V_2$ = 3 + $\frac{1.45\times T_2}{290}$
Again applying Charle's law,

$\therefore T_2=\frac{V_2{T}_1}{V_1}$
$=\frac{(3+\frac{1.45\times T_2}{290})\times 300}{3}=600K$

$\therefore T_2=600$ K = ${327}^{\circ }\text{C}$