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An insulated container of gas has two chambers separated by an insulating partition. One of the chambers has volume ${{V}_{1}}$ and contains ideal gas at pressure ${{p}_{1}}$ and temperature ${{T}_{1}}$. The other chamber has volume ${{V}_{2}}$ and contains ideal gas at pressure ${{p}_{2}}$ and temperature ${{T}_{2}}$. If the partition is removed without doing any work on the gas, the final equilibrium temperature of the gas in the container will be:
(a)- $\dfrac{{{T}_{1}}{{T}_{2}}({{p}_{1}}{{V}_{1}}+{{p}_{2}}{{V}_{2}})}{{{p}_{1}}{{V}_{1}}{{T}_{2}}+{{p}_{2}}{{V}_{2}}{{T}_{1}}}$
(b)- $\dfrac{{{p}_{1}}{{V}_{1}}{{T}_{1}}+{{p}_{2}}{{V}_{2}}{{T}_{2}}}{{{p}_{1}}{{V}_{1}}+{{p}_{2}}{{V}_{2}}}$
(c)- $\dfrac{{{p}_{1}}{{V}_{1}}{{T}_{2}}+{{p}_{2}}{{V}_{2}}{{T}_{1}}}{{{p}_{1}}{{V}_{1}}+{{p}_{2}}{{V}_{2}}}$
(d)- $\dfrac{{{T}_{1}}{{T}_{2}}({{p}_{1}}{{V}_{1}}+{{p}_{2}}{{V}_{2}})}{{{p}_{1}}{{V}_{1}}{{T}_{1}}+{{p}_{2}}{{V}_{2}}{{T}_{2}}}$

Last updated date: 13th Jun 2024
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Hint: It can be solved by using the formula of $\Delta U={{C}_{v}}\Delta T$ and applying the conservation of energy. The other equation that is used in this question is the ideal gas equation, i.e., $PV=nRT$.

Complete answer:
According to the law of conservation of mass, the energy before the equilibrium will be equal to energy after the equilibrium, which means that the internal energy before the equilibrium is equal to the internal energy after the equilibrium.
We know that:
$\Delta U={{C}_{v}}\Delta T$
Where ${{C}_{v}}$ is heat capacity at constant volume, T is the temperature, and U is the internal energy. So we can write it as:
Where T is the temperature after the equilibrium.
Suppose ${{n}_{1}}\text{ and }{{n}_{2}}$ are the moles of gases in each container. We can write:
Since the gases are ideal, we can write the equation:
Or we can write:
$(\dfrac{{{p}_{1}}{{V}_{1}}}{R{{T}_{1}}}+\dfrac{{{p}_{2}}{{V}_{2}}}{R{{T}_{2}}}){{C}_{v}}T=\left( \dfrac{{{p}_{1}}{{V}_{1}}}{R{{T}_{1}}} \right){{C}_{v}}{{T}_{1}}+\left( \dfrac{{{p}_{2}}{{V}_{2}}}{R{{T}_{2}}} \right){{C}_{v}}{{T}_{2}}$
$\left( \dfrac{{{p}_{1}}{{V}_{1}}}{R{{T}_{1}}}+\dfrac{{{p}_{2}}{{V}_{2}}}{R{{T}_{2}}} \right)T=\left( \dfrac{{{p}_{1}}{{V}_{1}}}{R} \right)+\left( \dfrac{{{p}_{2}}{{V}_{2}}}{R} \right)$
 $\left( \dfrac{{{p}_{1}}{{V}_{1}}{{T}_{2}}+{{p}_{2}}{{V}_{2}}{{T}_{1}}}{{{T}_{1}}{{T}_{2}}} \right)T={{p}_{1}}{{V}_{1}}+{{p}_{2}}{{V}_{2}}$
From this equation we can find the value of T:

Therefore, the correct answer is an option (a)- $\dfrac{{{T}_{1}}{{T}_{2}}({{p}_{1}}{{V}_{1}}+{{p}_{2}}{{V}_{2}})}{{{p}_{1}}{{V}_{1}}{{T}_{2}}+{{p}_{2}}{{V}_{2}}{{T}_{1}}}$.

The equation, $pV=nRT$can only be used when the gas is ideal, and this equation cannot be used if the gas is real. The equation of heat capacity at constant volume is taken because the system is in an insulated container.