Question

The density of a gas at ${27}^{\circ }C$ and 1 $atm$ is $d$. Pressure remaining constant, at which of the following temperature will its density become $0.75d$?

• A. ${20}^{\circ }C$
• B. ${30}^{\circ }C$
• C. 400 $K$
• D. 300 $K$

Answer 1

The correct option is C 400 $K$

Given that, $\left(d_1\right)=d$
$\left(d_2\right)=0.75d$

$P_1=1atm;T_1={27}^{\circ }C=300K$
$P_2=1atm;T_2=?$

$\left[\frac{d_1}{d_2}\right]=\frac{P_1\times T_2}{T_1\times P_2}$ or $\frac{T_2}{T_1}$

By putting all those values, we get-

$\therefore T_2=\frac{d_1}{d_2}\times T_1=\frac{d}{0.75d}\times 300=400K$

So, the correct option is $C$