Question

Two identical containers joined by a small pipe initially contain the same gas at pressure ${{P}_{0}}$​ and absolute temperature ${{T}_{0}}$. ​ One container is now maintained at the same temperature while the other is heated to $2{{P}_{0}}$​. The common pressure of the gasses will be-
A $\dfrac{3}{2}{{P}_{0}}$
B $\dfrac{4}{3}{{P}_{0}}$
C $\dfrac{5}{3}{{P}_{0}}$
D $2{{P}_{0}}$

Answer 1

Since there is no addition or removal of gas from the containers, the mass in the entire system will remain constant. Hence the number of molecules in the initial condition is the same as that of the number of molecules in the final stage. We have an ideal gas equation which relates volume, temperature, no of moles and pressure in a gaseous system. By using the equation we can find the pressure of the system when the 2nd container is heated up.

Formula used:
Conservation of mass,
$Initial\text{ }no.\text{ }of\text{ }moles\text{ }=\text{ }final\text{ }no.\text{ }of\text{ }moles$
$PV=\text{ }nRT$

Complete answer:
Since there is no additional quantity of gas added or removed from the system, mass remains the same. Then the number of moles will be the same.
i.e.,
If ${{n}_{1}}$ is the number of moles in 1st container and, ${{n}_{2}}$ is the number of moles in the second container,
$Initial\text{ }no.\text{ }of\text{ }moles\text{ }=\text{ }final\text{ }no.\text{ }of\text{ }moles$
Then,
${{\left( {{n}_{1}}+{{n}_{2}} \right)}_{i}}={{\left( {{n}_{1}}+{{n}_{2}} \right)}_{f}}$ ------ 1
Where,
${{\left( {{\text{n}}_{\text{1}}}\text{+}{{\text{n}}_{\text{2}}} \right)}_{\text{i}}}\text{= initial no of moles in the system }$
${{\left( {{\text{n}}_{\text{1}}}\text{+}{{\text{n}}_{\text{2}}} \right)}_{f}}\text{= final no of moles in the system}$
${{\text{n}}_{\text{1}}}\text{= no of moles in 1st container}$
${{n}_{2}}=\text{no of moles in 2nd container}$
To find pressure, we have ideal gas equation,
$PV=\text{ }nRT$
Where,
$\text{P = pressure }\left( \text{atm} \right)$
$\text{V= Volume}\left( \text{L} \right)$
$\text{n = no of moles}\left( \text{mol} \right)$
$\text{T = temperature}\left( \text{K} \right)$
$\text{R = universal gas constant = 8}\text{.314 J/mol}$
Then,
$n=\dfrac{PV}{RT}$ ------ 2
Substitute 2 in equation 1.
${{\left( \dfrac{{{P}_{1}}{{V}_{1}}}{R{{T}_{1}}}+\dfrac{{{P}_{2}}{{V}_{2}}}{R{{T}_{2}}} \right)}_{i}}={{\left( \dfrac{{{P}_{1}}{{V}_{1}}}{R{{T}_{1}}}+\dfrac{{{P}_{2}}{{V}_{2}}}{R{{T}_{2}}} \right)}_{f}}$ ---------- 3
Where,
${{\text{P}}_{\text{1}}}\text{ and }{{\text{P}}_{\text{2}}}\text{= Pressure in containers 1 and 2}$
${{\text{T}}_{\text{1}}}\text{ and }{{\text{T}}_{\text{2}}}\text{= Temperature in containers 1 and 2}$
${{V}_{\text{1}}}\text{ and }{{\text{V}}_{\text{2}}}\text{= Volume in containers 1 and 2}$
Initially we have,
${{\text{P}}_{1}}\text{ = }{{\text{P}}_{2}}\text{= }{{P}_{0}}$
${{T}_{1}}={{T}_{2}}={{T}_{0}}$
${{V}_{1}}={{V}_{2}}=V$
When 2nd container is heated,
${{T}_{1}}={{T}_{0}}$
${{T}_{2}}=2{{T}_{0}}\text{ }$
${{\text{V}}_{1}}={{V}_{2}}$
We need to find ${{P}_{1}}$ and ${{P}_{2}}$ when the 2nd container is heated.
Let’s take, ${{P}_{1}}={{P}_{2}}=P$and ${{V}_{1}}={{V}_{2}}=V$
Substitute the values of ${{\text{V}}_{1}},{{\text{V}}_{2}},{{P}_{1}},{{P}_{2}},{{T}_{1}},{{T}_{2}}$ In equation 3, we get,
$\dfrac{{{P}_{0}}V}{R{{T}_{0}}}+\dfrac{{{P}_{0}}V}{R{{T}_{0}}}=\dfrac{PV}{R{{T}_{0}}}+\dfrac{PV}{2{{T}_{0}}R}$
By solving the above equation,
We get,
$P=\dfrac{4}{3}{{P}_{0}}$

So, the correct answer is “Option B”.

Note:
The Ideal Gas Equation is used to determine the relationship between the pressure, volume, no of moles and temperature of an ideal gas in a defined environment which will be controlled for constant volume. It can be used to determine the density of a gas sample for a given pressure and temperature.