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# A gas in an open container is heated from $27^{\large\circ}C$ to $127^{\large\circ}C$. The fraction of the original amount of gas remaining in the container will be

$(a)\;\large\frac{3}{4}\qquad(b)\;\large\frac{1}{4}\qquad(c)\;\large\frac{1}{2}\qquad(d)\;\large\frac{1}{8}$

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A)
On heating a gas in a open container (P = constant, V = Constant) some moles of gas goes out.
Thus,
$n_1T_1 = n_2T_2$
$n_1\times300 = n_2\times400$
$n_2 = \large\frac{3n_1}{4}$
$\therefore$ Fraction of moles of gas left
$\Rightarrow \large\frac{n_2}{n_1} = \large\frac{3n_2}{4n_1} =\large\frac{3}{4}$
Hence answer is (a)
The last step should be like this,

$\frac{3n_1}{4n_1} = \frac{3}{4}$