Question

A gas in an open container is heated from ${27}^{{}^{\circ }}C$ to ${127}^{{}^{\circ }}C$, the fraction of the original of gas remaining in the container will be:

• A. $\frac{3}{4}$
• B. $\frac{1}{4}$
• C. $\frac{1}{2}$
• D. $\frac{1}{8}$

Answer 1

Answer: (A)

$\frac{3}{4}$

We know that, $PV=nRT=$ constant

Where, Pressure P and Volume V are constant, in this case.

n is number of moles

R is gas constant

T is temperature

Given two temperature are 27 and 127 degree celcius

$T_1=27+273=300K$

$T_2=127+273=400K$

Therefore, $n_1{T}_1=n_2{T}_2$

$n_1\times 300=n_2\times 400$

$n_2=\frac{3n_1}{4}$

So, the fraction of moles of the remaining gas in the container after heating will be $\frac{n_2}{n_1}=\frac{3}{4}$

Hence, option $A$ is correct.