Question

The total kinetic energy of translatory motion of all molecules of 5 litres of nitrogen exerting a pressure is 3000J.
This question has multiple correct options
A. The total K.E. of 10 litres of ${N}_{2}$ at pressure of 2P is 3000 J.
B. The total K.E. of 10 litres of He at pressure of 2P is 3000 J.
C. The total K.E. of 10 litres of ${O}_{2}$ at pressure of 2P is 20000 J.
D. The total K.E. of 10 litres of Ne at pressure of 2P is 12000 J.

Answer 1

Hint: To solve this question, use the formula for total kinetic energy of a gas in terms of pressure and volume. Substitute the value of energy and volume given in the question in this formula and calculate the value of P i.e. pressure. Then, use this same formula, substitute the values in the formula and cross-check every option by substituting the values given in that specific option. It is given that this question has multiple correct options, so take care of it.

Formula used:
$E= \dfrac {f}{2}PV$

Complete step-by-step solution:
Given:
V= 5 litres
E= 3000J
Total energy of a gas is given by,
$E= \dfrac {f}{2}PV$ …(1)
Where,
f is the degree of freedom
P is the pressure
V is the volume of gas

Substituting values in the equation. (1) gives total kinetic energy of translatory motion of 5 litres of nitrogen,
$3000= \dfrac {f}{2} \times P \times 5 $
We know, f=3 for translatory motion
$\therefore 3000=\dfrac {3}{2}\times P \times 5 $
$\Rightarrow P= \dfrac {3000 \times 2}{3 \times 5 }$
$\Rightarrow P= 400 {N}/{{m}^{2}}$ …(2)

A) The total kinetic energy of 10 litres of ${N}_{2}$ at pressure of 2P can be calculated by substituting value in equation. (1),
$E= \dfrac {f}{2} \times 2P \times 10$
Substituting equation. (2) in above equation we get,
$E= \dfrac {f}{2} \times \left (2 \times 400 \right) \times 10 $…(3)
We know, ${N}_{2}$ is a diatomic molecule and degrees of freedom for a diatomic molecule is 5.
$\therefore f=5$
Substituting value of f in equation. (3) we get,
$E= \dfrac {5}{2} \times \left (2 \times 400 \right) \times 10$
$\Rightarrow E= 20000 J$

B) The total kinetic energy of 10 litres of He at pressure of 2P can be calculated by substituting value in equation. (1),
$E= \dfrac {f}{2} \times 2P \times 10$
Substituting equation. (2) in above equation we get,
$E= \dfrac {f}{2} \times \left (2 \times 400 \right) \times 10$ …(4)
We know, He is a monoatomic molecule and degrees of freedom for a monoatomic molecule is 3.
$\therefore f=3$
Substituting value of f in equation. (4) we get,
$E= \dfrac {3}{2} \times \left (2 \times 400 \right) \times 10 $
$\Rightarrow E= 12000 J$

C) The total kinetic energy of 10 litres of ${O}_{2}$ at pressure of 2P can be calculated by substituting value in equation. (1),
$E= \dfrac {f}{2} \times 2P \times 10$
Substituting equation. (2) in above equation we get,
$E= \dfrac {f}{2} \times \left (2 \times 400 \right) \times 10 $ …(5)
We know, ${O}_{2}$ is a diatomic molecule and degrees of freedom for a diatomic molecule is 5.
$\therefore f=5$
Substituting value of f in equation. (5) we get,
$E= \dfrac {5}{2} \times \left (2 \times 400 \right) \times 10$
$\Rightarrow E= 20000 J$

D) The total kinetic energy of 10 litres of Ne at pressure of 2P can be calculated by substituting value in equation. (1),
$E= \dfrac {f}{2} \times 2P \times 10$
Substituting equation. (2) in above equation we get,
$E= \dfrac {f}{2} \times \left (2 \times 400 \right) \times 10 $ …(6)
We know, He is a monoatomic molecule and degrees of freedom for a monoatomic molecule is 3.
$\therefore f=3$
Substituting value of f in equation. (6) we get,
$E= \dfrac {3}{2} \times \left (2 \times 400 \right) \times 10 $
$\Rightarrow E= 12000 J$

So, the correct answers are option C and D.

Note:
It is given that this question has multiple correct options, so students must take care of it. To solve these types of questions, students must remember the degrees of freedom of monoatomic, diatomic and triatomic gases. For monoatomic gases, degrees of freedom (f) is 3; for diatomic gases, it is f=5; for triatomic gases, f=6. But, these degrees of freedom are correct only if there is no vibrational motion. If there’s a vibrational motion then these degrees of freedom vary.