Question

The temperature of a gas placed in an open container is raised from ${27}^{o}C$ to ${227}^{o}C$. The percent of the original amount of the gas expelled from the container will be:

• A. $20$
• B. $40$
• C. $60$
• D. $80$

Answer 1

Answer: (B)

$40$

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,

$n_1=n$

$n_2=?$

$T_1=27℃=(27+273)K=300K$

$T_2=227℃=(227+273)K=500K$

$\therefore n_2=\frac{n\times 300}{500}$

$\Rightarrow n_2=0.6n$

$\therefore$ Amount of gas expelled = original amount of gas - amount of gas left $=n_1-n_2=n-0.6n=0.4n$

Percent of original amount of the gas expelled $=\frac{\text{amount of gas expelled}}{\text{original amount of gas}}\times 100=\frac{0.4n}{n}\times 100=40\%$

Hence, $40\%$ of the original amount of the gas expelled from the container.