Question

If $ 2moles $ of an ideal monatomic gas at a temperature $ {T_0} $ is mixed with $ 4mole $ of another ideal monatomic gas at a temperature $ 2{T_0} $ then the temperature of the mixture is:
(A) $ \dfrac{5}{3}{T_0} $
(B) $ \dfrac{3}{2}{T_0} $
(C) $ \dfrac{4}{3}{T_0} $
(D) $ \dfrac{5}{4}{T_0} $

Answer 1

Hint: Use the energy conservation law and law of equipartition energy to solve the problem. Law of conservation of energy: $ E = {E_1} + {E_2} $ where, $ {E_1} $ and $ {E_2} $ are the individual internal energies of the components of mixture respectively and $ E $ is the internal energy of the total mixture.
Law of Equipartition of Energy: $ E = \dfrac{1}{2}fnRT $ where $ f $ is the degree of freedom, $ n $ is the number of moles of the ideal gas, $ R $ is the universal gas constant and $ T $ is the absolute temperature .

Complete step by step solution:
From the given question we have the information about the number of moles of the ideal gases, their temperature and their type i.e., monoatomic gases. We are asked to calculate the temperature of the mixture of the two gases.
We know by conservation of energy that total internal energy of a system remains conserved. So, if $ {E_1} $ be the internal energy of the monatomic gas at temperature $ {T_0} $ , $ {E_2} $ be the internal energy of the monatomic gas at temperature $ 2{T_0} $ and $ E $ be the internal energy of the total mixture then by law of conservation of energy we can write:
 $ E = {E_1} + {E_2} $ ............................... $ \to eqn(1) $
The law of equipartition of energy states that for a system of particles at absolute temperature $ T $ ,the average internal energy associated with each degree of freedom( $ f $ ) is $ \dfrac{1}{2}fnRT $ where $ n $ and $ R $ have the same physical meanings as stated above.
Thus, substituting the value of internal energy in $ eqn(1) $ we can write:
 $ \dfrac{1}{2}f({n_1} + {n_2})RT = \dfrac{1}{2}f{n_1}R{T_1} + \dfrac{1}{2}f{n_2}R{T_2} $ $ \to eqn(2) $
Where, $ {n_1} = 2mole $ ; $ {n_2} = 4mole $ ; $ {T_1} = {T_0} $ ; $ {T_2} = 2{T_0} $ (from the given question)
Cancelling $ f $ and $ R $ on both sides as the degrees of freedom is equal for all monatomic gases and Universal gas constant is a constant term for all gases, we can rewrite $ eqn(2) $ as:
    $ ({n_1} + {n_2})T = {n_1}{T_1} + {n_2}{T_2} $
Substituting the values, we have,
 $ \Rightarrow (2 + 4)T = 2{T_0} + 4(2{T_0}) $
 $ \Rightarrow T = \dfrac{{{\text{10}}}}{6}{T_0} = \dfrac{5}{3}{T_0} $
Thus, option A is the correct answer to the question.

Note:
Degrees of freedom can be defined as the minimum number of coordinates required to specify the configuration of a dynamic system.
For a monatomic gas $ f = 3 $ .