Question 634: Thermodynamics & KTG

Question

500 gram of a diatomic gas is enclosed at a pressure of $10^5 \mathrm{Nm}^{-2}$. The density of the gas is $5 \mathrm{kgm}^{-3}$. The energy of one mole of the gas due to its thermal motion is [consider the gas molecule as a rigid rotator]

Accepted Answer

Energy of one mole of gas due to thermal motion means kinetic energy of the gas per mole, which is given by
$$ \begin{aligned} & 	ext { K.E. }=\frac{f}{2} R T=\frac{f}{2} P V \ & V=\frac{	ext { Mass }}{	ext { Density }}=\frac{500 	imes 10^{-3}}{5}=10^{-1} \mathrm{~m}^3 \ & 	ext { K.E. }=\frac{f}{2} P V \quad \ldots .(f=5 	ext { for diatomic gas }) \ & 	ext { K.E. }=\frac{5}{2} P V=\frac{5}{2} 	imes 10^5 	imes 10^{-1}=2.5 	imes 10^4 \mathrm{~J} \end{aligned} $$

Answer

$1.5 	imes 10^4 \mathrm{~J}$

Answer

$2.5 	imes 10^4 \mathrm{~J}$

Answer

$1.5 	imes 10^7 \mathrm{~J}$

Answer

$2.5 	imes 10^7 \mathrm{~J}$

Thermodynamics & KTG

500 gram of a diatomic gas is enclosed at a pressure of 10^5 \mathrm{Nm}^{-2}. The density of the gas is 5 \mathrm{kgm}^{-3}. The energy of one mole of the gas due to its thermal motion is [consider the gas molecule as a rigid rotator]

Choose the correct answer: