Question

Two monatomic ideal gas at temperature ${T}_{1}$ and ${T}_{2}$ are mixed. There is no loss of energy. If the masses of molecules of the gases are ${m}_{1}$ and ${m}_{2}$ and the number of their molecules are ${n}_{1}$ and ${n}_{2}$ respectively. The temperature of the mixture will be:
A. $\dfrac {{T}_{1}+{T}_{2}}{{n}_{1}+{n}_{2}}$
B. $\dfrac {{T}_{1}}{{n}_{1}}+\dfrac {{T}_{2}}{{n}_{2}}$
C. $\dfrac {{n}_{2}{T}_{1}+{n}_{1}{T}_{2}}{{n}_{1}+{n}_{2}}$
D. $\dfrac {{n}_{1}{T}_{1}+{n}_{2}{T}_{2}}{{n}_{1}+{n}_{2}}$

Answer 1

Use the formula for internal energy and find the internal energy at temperature ${T}_{1}$ and then at ${T}_{2}$. As there is no loss of energy, the sum of change of internal energies must be zero. Substitute the values in the mentioned expression. Now, substitute the value for internal energy of gas at constant volume. Then, consider T as the final temperature of the mixture, substitute it in the above obtained expression. Rearrange the equation and obtain the temperature of the mixture.
Formula used:
$\Delta U=n{C}_{v} \Delta T$

Complete answer:
It is given that there is no loss of energy which means that the sum of change of internal energies must be zero. Mathematically this can be written as,
$\Delta {U}_{1}+ \Delta {U}_{2}=0$ …(1)
Where, $\Delta {U}_{1}$ is the internal energy at temperature ${T}_{1}$
$\Delta {U}_{2}$ is the internal energy at temperature ${T}_{2}$
We know, internal energy is given by,
$\Delta U=n{C}_{v} \Delta T$
Where, n is the number of molecules
${C}_{v}$ is the internal energy of gas at constant volume
Using above equation we can write the internal energy at temperature ${T}_{1}$ as,
$\Delta {U}_{1}={n}_{1}{C}_{{v}_{1}} \Delta {T}_{1}$ …(2)
Similarly, we can write the internal energy at temperature ${T}_{2}$ as,
$\Delta {U}_{2}={n}_{2}{C}_{{v}_{2}} \Delta {T}_{2}$ …(2)
Substituting equation. (2) and (3) in equation. (1) we get,
${n}_{1}{C}_{{v}_{1}} \Delta {T}_{1}+ {n}_{2}{C}_{{v}_{2}} \Delta {T}_{2}$ …(4)
We know, for monoatomic gases,
${C}_{v}= \dfrac {3}{2}R$
It is given that both the gases are monatomic. Thus,
${C}_{{v}_{1}}={C}_{{v}_{2}}= \dfrac {3}{2}R$
Substituting this value in the equation. (4) we get,
$\dfrac {3}{2}R{n}_{1} \Delta {T}_{1}+ \dfrac {3}{2}R{n}_{2} \Delta {T}_{2}=0$
$\therefore \dfrac {3}{2}R \left ({n}_{1}\Delta {T}_{1}+{n}_{2} \Delta {T}_{2} \right )=0$ …(5)
Let the final temperature be T.
Thus, equation. (5) can be written as,
$\dfrac {3}{2}R \left ({n}_{1}\left(T-\Delta {T}_{1} \right)+{n}_{2} \left (T-\Delta {T}_{2}\right) \right )=0$
Rearranging the above equation we get,
$T=\dfrac {{n}_{1}{T}_{1}+{n}_{2}{T}_{2}}{{n}_{1}+{n}_{2}}$
Thus, the temperature of the mixture will be $\dfrac {{n}_{1}{T}_{1}+{n}_{2}{T}_{2}}{{n}_{1}+{n}_{2}}$.

So, the correct answer is option D i.e. $\dfrac {{n}_{1}{T}_{1}+{n}_{2}{T}_{2}}{{n}_{1}+{n}_{2}}$.

Note:
To solve these types of questions, students must have the clear understanding of thermodynamics and its laws. Also, they should know the ideal gas law and how varying the different parameters changes the behaviour of gases at various conditions.