Question

Two rigid boxes containing two different ideal gases are placed on table. Box A contains 1 mole of nitrogen at temperature ${T_0}$, while box B contains 1 mole of helium at temperature $7/3{T_0}$. The boxes are then put into the thermal contact with each other and heat flows between them until the gas reach a common final temperature (ignore the heat capacity of boxes) then the final temperature ${T_f}$ of the gases, in terms of ${T_0}$ is
A. $\dfrac{{2{T_0}}}{5}$
B. $\dfrac{{3{T_0}}}{2}$
C. $\dfrac{{5{T_0}}}{3}$
D. $\dfrac{{9{T_0}}}{7}$

Answer 1

Use the formula for the change in internal energy of a gas. This formula gives the relation between the number of moles of the gas, specific heat of the gas and change in temperature of the gas. As the two boxes are kept in thermal equilibrium, they exchange heat with each other. But the change in the internal energy of the system of two gases is zero.

Formulae used:
The formula for change in internal energy $\Delta U$ of an ideal gas is
$\Rightarrow \Delta U = n{C_V}\Delta T$ …… (1)
Here, $n$ is the number of moles of gas, ${C_V}$ is the specific heat of the gas and $\Delta T$ is the change in temperature of the gas.
The specific heat ${C_V}$ for monatomic gas is
$\Rightarrow {C_V} = \dfrac{3}{2}R$ …… (2)
Here, $R$ is the gas constant.
The specific heat ${C_V}$ for diatomic gas is
$\Rightarrow {C_V} = \dfrac{5}{2}R$ …… (3)
Here, $R$ is the gas constant.

Complete step by step solution:
We have given that there are two boxes placed on a table containing 1 mole of nitrogen and 1 mole of helium is each.
The temperature of the nitrogen gas is ${T_0}$ and the temperature of the helium gas is $\dfrac{7}{3}{T_0}$.
Rewrite equation (1) for the nitrogen gas.
$\Rightarrow \Delta {U_{nitrogen}} = {n_{nitrogen}}{C_{{V_{nitrogen}}}}\Delta {T_{nitrogen}}$
Nitrogen always exists as a diatomic gas.
Substitute $1\,{\text{mol}}$ for ${n_{nitrogen}}$, $\dfrac{5}{2}R$ for ${C_{{V_{nitrogen}}}}$ and ${T_f} - {T_0}$ for $\Delta {T_{nitrogen}}$ in the above equation.
$\Rightarrow \Delta {U_{nitrogen}} = \left( {1\,{\text{mol}}} \right)\left( {\dfrac{5}{2}R} \right)\left( {{T_f} - {T_0}} \right)$
$\Rightarrow \Delta {U_{nitrogen}} = \dfrac{5}{2}R\left( {{T_f} - {T_0}} \right)$
Here, ${T_f}$ is the final temperature attained by nitrogen when kept in thermal contact with helium.
Rewrite equation (1) for the helium gas.
$\Rightarrow \Delta {U_{helium}} = {n_{helium}}{C_{{V_{helium}}}}\Delta {T_{helium}}$
Nitrogen always exists as a monatomic gas.
Substitute $1\,{\text{mol}}$ for ${n_{helium}}$, $\dfrac{3}{2}R$ for ${C_{{V_{helium}}}}$ and ${T_f} - \dfrac{7}{3}{T_0}$ for $\Delta {T_{helium}}$ in the above equation.
$\Rightarrow \Delta {U_{helium}} = \left( {1\,{\text{mol}}} \right)\left( {\dfrac{3}{2}R} \right)\left( {{T_f} - \dfrac{7}{3}{T_0}} \right)$
$\Rightarrow \Delta {U_{helium}} = \dfrac{3}{2}R\left( {{T_f} - \dfrac{7}{3}{T_0}} \right)$
Here, ${T_f}$ is the final temperature attained by helium when kept in thermal contact with nitrogen.
When the two boxes are in thermal contact with each other, the gases in the two boxes exchange heat with each other but not with the surrounding.
Hence, the change in the internal energy of the system of two gases is zero.
$\Rightarrow \Delta {U_{nitrogen}} + \Delta {U_{helium}} = 0$
Substitute $\dfrac{5}{2}R\left( {{T_f} - {T_0}} \right)$ for $\Delta {U_{nitrogen}}$ and $\dfrac{3}{2}R\left( {{T_f} - \dfrac{7}{3}{T_0}} \right)$ for $\Delta {U_{helium}}$ in the above equation.
$\Rightarrow \dfrac{5}{2}R\left( {{T_f} - {T_0}} \right) + \dfrac{3}{2}R\left( {{T_f} - \dfrac{7}{3}{T_0}} \right) = 0$
$\Rightarrow \dfrac{5}{2}R{T_f} - \dfrac{5}{2}R{T_0} + \dfrac{3}{2}R{T_f} - \dfrac{3}{2}R{T_f}\dfrac{7}{3}{T_0} = 0$
$\Rightarrow 4R{T_f} - \dfrac{5}{2}R{T_0} - \dfrac{{21}}{6}R{T_0} = 0$
$\Rightarrow 4R{T_f} - \left( {\dfrac{{30 + 42}}{{12}}} \right)R{T_0} = 0$
$\Rightarrow 4R{T_f} - 6R{T_0} = 0$
$\Rightarrow 2{T_f} = 3{T_0}$
$\therefore {T_f} = \dfrac{{3{T_0}}}{2}$
Therefore, the final temperature of the gas will be $\dfrac{{3{T_0}}}{2}$.

Hence, the correct option is B.

Note: One can also solve the same question in another way. From the information given in the question, we can conclude that the initial temperature of the nitrogen gas is less than the initial temperature of helium gas. Hence, equate the loss in internal energy of the helium gas to the gain in energy of the nitrogen gas and determine the final temperature of the gas.