Question

An insulated container of gas has two chambers separated by an insulating partition. One of the chambers has volume ${V_{1}}$ and contains an ideal gas at pressure ${P_{1}}$ and temperature ${T_{1}}$. The other chamber has volume ${V_{2}}$ and contains the same 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, the final equilibrium temperature of the gas in the container will be:
A.$

$$\begin{aligned} \dfrac{T_{1}T_{2}(P_{1}V_{1}+P_{2}V_{2})}{P_{1}V_{1}T_{2}+P_{2}V_{2}T_{1}} \end{aligned}$$

$B.$

$$\begin{aligned} \dfrac{P_{1}V_{1}T_{1}+P_{2}V_{2}T_{2}}{P_{1}V_{1}+P_{2}V_{2}} \end{aligned}$$

$C.$

$$\begin{aligned} \dfrac{P_{1}V_{1}T_{2}+P_{2}V_{2}T_{1}}{P_{1}V_{1}+P_{2}V_{2}} \end{aligned}$$

$D.$

$$\begin{aligned} \dfrac{T_{1}T_{2}(P_{1}V_{1}+P_{2}V_{2})}{P_{1}V_{1}T_{1}+P_{2}V_{2}T_{2}} \end{aligned}$$

$

Answer 1

It can be solved by using the formula of $

$$\begin{aligned} \Delta U=C_{v}\Delta T \end{aligned}$$

$and applying the conservation of energy. The other equation that is used in this question is the ideal gas equation, i.e.,$

$$\begin{aligned} PV=nRT \end{aligned}$$

$

Complete step-by-step answer: According to the law of conservation of mass, the energy before the equilibrium will be equal to the energy after equilibrium, which means that the internal energy before the equilibrium is equal to the internal energy after the equilibrium.
We know that
$

$$\begin{aligned} \Delta U=C_{v}\Delta T \end{aligned}$$

$Where, C_{v}= heat capacity at constant volume T = temperature U = internal energy So,$

$$\begin{aligned} C_{v}T=C_{v}T_{1}+C_{v}T_{2} \end{aligned}$$

$Where, T is the temperature after equilibrium. Let us assume${n_{1}}$and${n_{2}}$are the moles of gases in each container. So,$

$$\begin{aligned} (n_{1}+n_{2})C_{v}T=n_{1}C_{v}T_{1}+n_{2}C_{v}T_{2} \end{aligned}$$

$Since the gases are ideal, we can use the equation$

$$\begin{aligned} PV=nRT \end{aligned}$$

$$

$$\begin{aligned} n=\dfrac{pV}{RT} \end{aligned}$$

$Or we can write,$

$$\begin{aligned} \left( \dfrac{p_{1}V_{1}}{RT_{1}}+\dfrac{p_{2}V_{2}}{RT_{2}} \right) T=\left( \dfrac{p_{1}V_{1}}{R} \right) +\left( \dfrac{p_{2}V_{2}}{R} \right) \end{aligned}$$

$From this equation, we will get the value of T as:$

$$\begin{aligned} \dfrac{T_{1}T_{2}(P_{1}V_{1}+P_{2}V_{2})}{P_{1}V_{1}T_{2}+P_{2}V_{2}T_{1}} \end{aligned}$$

$

Therefore, the correct option is (A).

Note: The ideal gas equation 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.