Question

A gas in an open container is treated from 2$7^{o}C$ to ${127}^{0}C$. The fraction of original amount 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

The correct option is A $\frac{3}{4}$

As we know that $PV=nRT$. Here pressure P and volume V are kept constant.

Here $T_1$ = 300 K and $T_2$ = 400 K

Let $n_1,n_2$ are the mole numbers at temp $T_1$ and $T_2$

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

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

So,$n_2=3n_1/4$.

So the correct option is A