Question

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

Answer 1

Hint: To answer this question, we have to use the formula for kinetic energy of a molecule of the gas. Then, we have to manipulate this formula in terms of the given parameters to get the final answer.

Formula used:
The formulae used to solve this question are given by
$\Rightarrow E = \dfrac{3}{2}{k_B}T$
$\Rightarrow {k_B} = \dfrac{R}{{{N_A}}}$
$\Rightarrow PV = nRT$
Here $E$ is the kinetic energy of a gas molecule at an absolute temperature of $T$, ${k_B}$ is the Boltzmann’s constant, ${N_A}$ is the Avogadro’s constant,$R$ is the universal gas constant, $n$ is the number of moles.

Complete step by step solution:
We know that the kinetic energy of a molecule of gas due to its thermal motion is given by
$\Rightarrow E = \dfrac{3}{2}{k_B}T$
If the given gas has $N$ molecules, then the total kinetic energy of the gas is
$\Rightarrow {E_T} = NE$
$\Rightarrow {E_T} = \dfrac{{3N}}{2}{k_B}T$
We know that ${k_B} = \dfrac{R}{{{N_A}}}$
So that
$\Rightarrow {E_T} = \dfrac{{3N}}{2}\dfrac{R}{{{N_A}}}T$
$\Rightarrow {E_T} = \dfrac{3}{2} \times R \times \dfrac{N}{{{N_A}}} \times T$
As we know from the Avogadro’s law that
$\Rightarrow N = n{N_A}$
Putting this in the above expression
$\Rightarrow {E_T} = \dfrac{3}{2} \times R \times \dfrac{{n{N_A}}}{{{N_A}}} \times T$
$\Rightarrow {E_T} = \dfrac{3}{2}nRT$ ………………………..(i)
From the ideal gas equation, we have
$\Rightarrow PV = nRT$
So (i) becomes
$\Rightarrow {E_T} = \dfrac{3}{2}PV$ ……………………….(ii)
As the density $\rho = \dfrac{M}{V}$
So $V = \dfrac{M}{\rho }$
Putting in (ii) we get
$\Rightarrow {E_T} = \dfrac{3}{2}\dfrac{{PM}}{\rho }$
Now, according to the question, we have the pressure $P = 8 \times {10^4}N/{m^2}$ , mass of the gas $M = 1kg$ and the density of the gas $\rho = 4kg/{m^3}$. Substituting these in the above expression, we get
 $\Rightarrow {E_T} = \dfrac{3}{2} \times \dfrac{{8 \times {{10}^4} \times 1}}{4}$
On solving we finally get
$\Rightarrow {E_T} = 3 \times {10^4}J$
Thus, the energy of the gas due to the thermal motion is equal to $3 \times {10^4}J$.
Hence, the correct answer is option (A).

Note:
Don’t get confused by the atomicity of the gas given in the question. We have derived the final expression for the total kinetic energy of the gas starting from the very basic formula for the energy of a gas molecule, which is the same for all the gases regardless of its atomicity.