Q. The specific heats of an ideal gas at constant pressure and constant volume are 525 J $(kg{^\circ}C)^{- 1}$ and $315\ J\ (kg{^\circ}\ C)^{- 1}$ respectively. Its density at NTP is
A) $0.64\ kgm^{- 3}$
B) $1.20\ kgm^{- 3}$
C) $1.75\ kgm^{- 3}$
D) $2.62\ kgm^{- 3}$
Explanation
(c) If $M$ is molecular mass of the gas, then from
$M\left( C_{p} - C_{v} \right) = R$
$M = \frac{8.31}{210} = 0.0392$
If $\rho$ is density of the gas at NTP, then mass of 1 $m^{3}$of gas at NTP$= \rho$ kg
$\therefore$ Mass of 22.4 L $( = 22.4 \times 10^{- 3}m^{3})$ of gas at NTP $= \rho \times 22.4 \times 10^{- 3}\ $kg, which is the molecular mass of the gas
$\therefore\ \rho \times 22.4 \times 10^{- 3} = 0.0392$
$\rho = \frac{0.0392}{22.4 \times 10^{- 3}} = 1.75\ kgm^{- 3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.