Q. The pressure and temperature of $4{dm}^{3}$ of carbon dioxide gas are doubled, then the volume of carbon dioxide gas would be
A) $2\ {dm}^{3}$
B) $3\ {dm}^{3}$
C) $4\ {dm}^{3}$
D) $8\ {dm}^{3}$
Explanation
(c)
Given, $\frac{p_{2}}{p_{1}} = 2,\frac{T_{2}}{T_{1}} = 2,\ V_{1} = 4\ {dm}^{3},\ V_{2} = ?$
From gas equation
$\frac{p_{1}V_{1}}{T_{1}} = \frac{p_{2}V_{2}}{T_{2}}$
$or\ \frac{V_{1}}{V_{2}} = \frac{p_{2}}{p_{1}} \times T_{1}/T_{2}$
$\therefore\ \frac{4}{V_{2}} = 2 \times \frac{1}{2} = 1$
$\therefore\ V_{2} = 4\ {dm}^{3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.