Question

One kg of a diatomic gas is at a pressure of $8 \times {10^4}\,{\text{N/}}{{\text{m}}^2}$. The density of the gas is $4\,{\text{kg/}}{{\text{m}}^3}$. The energy of the gas due to its thermal motion will be
A. $3 \times {10^4}\,{\text{J}}$
B. $5 \times {10^4}\,{\text{J}}$
C. $6 \times {10^4}\,{\text{J}}$
D. $7 \times {10^4}\,{\text{J}}$

Answer 1

Use the formula for the thermal energy of the diatomic gas. Also use the ideal gas equation. Determine the volume of the gas using the formula for density of the gas. Derive the relation between thermal energy of gas, volume of gas and pressure of the gas.

Formula used:
The thermal energy $U$ of a diatomic gas is given by
$U = \dfrac{5}{2}nRT$ …… (1)
Here, $n$ is the number of moles of gas, $R$ is the gas constant and $T$ is the temperature of the gas.
The ideal gas law is given by
$PV = nRT$ …… (2)
Here, $P$ is the pressure of gas, $V$ is the volume of the gas, $n$ is the number of moles of gas, $R$ is the gas constant and $T$ is the temperature of the gas.
The density $\rho$ is given by
$\rho = \dfrac{M}{V}$ …… (3)
Here, $M$ is the mass and $V$ is the volume.

Complete step by step answer:
One kg of a diatomic gas is at a pressure of $8 \times {10^4}\,{\text{N/}}{{\text{m}}^2}$. The density of the gas is $4\,{\text{kg/}}{{\text{m}}^3}$.
$M = 1\,{\text{kg}}$
$P = 8 \times {10^4}\,{\text{N/}}{{\text{m}}^2}$
$\rho = 4\,{\text{kg/}}{{\text{m}}^3}$
Determine the volume of the diatomic gas.
Rearrange equation (3) for volume $V$ of the gas.
$V = \dfrac{M}{\rho }$
Substitute $1\,{\text{kg}}$ for $M$ and $4\,{\text{kg/}}{{\text{m}}^3}$ for $\rho$ in the above equation.
$V = \dfrac{{1\,{\text{kg}}}}{{4\,{\text{kg/}}{{\text{m}}^3}}}$
$\Rightarrow V = \dfrac{1}{4}\,{{\text{m}}^3}$
Hence, the volume of the gas is $\dfrac{1}{4}\,{{\text{m}}^3}$.
Determine the energy of the diatomic gas.
Substitute $PV$ for $nRT$ in equation (1).
$U = \dfrac{5}{2}PV$
Substitute $8 \times {10^4}\,{\text{N/}}{{\text{m}}^2}$ for $P$ and $\dfrac{1}{4}\,{{\text{m}}^3}$ for $V$ in the above equation.
$U = \dfrac{5}{2}\left( {8 \times {{10}^4}\,{\text{N/}}{{\text{m}}^2}} \right)\left( {\dfrac{1}{4}\,{{\text{m}}^3}} \right)$
$\Rightarrow U = 5 \times {10^4}\,{\text{J}}$

Therefore, the energy of the gas due to its thermal motion is $5 \times {10^4}\,{\text{J}}$.

So, the correct answer is “Option B”.

Note:
The thermal energy for the monatomic, diatomic, triatomic gases is not the same. These equations for thermal energy are different for different gases and depend on the number of atoms in the gas molecule. These equations vary with the number of atoms in the molecule because gases with different atoms have different degrees of freedom.