Question

A gas in an open container is heated from ${27}^{o}C$ to ${127}^{o}C$. The fraction of the original amounts of gas left in the container would be:

• A. $3/4$
• B. $1/2$
• C. $2/3$
• D. $7/8$

Answer 1

The correct option is A $3/4$

From ideal gas equation-

$PV=nRT$

whereas,

$P=$ pressure of the gas

$V=$ volume it occupies

$n=$ number of moles of gas present in the sample

$R=$ universal gas constant $=0.0821atmL{mol}^{-1}{K}^{-1}$

$T=$ absolute temperature of the gas

At constant pressure and volume,

$n\propto \frac{1}{T}$

$\therefore \frac{n_1}{n_2}=\frac{T_2}{T_1}$

$\Rightarrow n_2=\frac{n_1\times T_1}{T_2}$

Let the original amount of the gas be n moles.

Therefore,

$n_1=n$

$n_2=?$

$T_1=27℃=(27+273)K=300K$

$T_2=227℃=(127+273)K=400K$

$\therefore n_2=\frac{n\times 300}{400}$

$\Rightarrow n_2=\frac{3}{4}n$

Hence, the fraction of the original amounts of gas left in the container would be $\frac{3}{4}$.