A container is divided into two chambers by a partition. The volume of the first chamber is 4.5 litres and the second chamber is 5.5 litres. The first chamber contains 3.0 mol of gas at a pressure of 2.0 atm and the second chamber contains 4.0 mol of identical gas at a pressure of 3.0 atm. After the partition is removed and the mixture attains equilibrium, then, the common equilibrium pressure existing in the mixture is \[x \times {10^{ - 1}}atm\]. Find the value of x.

Hint: Before finding the solution to this problem, we need to know about the ideal gas equation or law. It states that the product of the pressure and the volume of one gram molecule of an ideal gas is equal to the product of the absolute temperature of the gas and the universal gas constant. That is, \[PV = nRT\]. Using this we are going to solve this problem.

Formula Used:
The formula for the ideal gas equation is,
\[PV = nRT\]
Where, P is pressure, V is volume, n is number of moles, R is universal gas constant and T is temperature.

Complete step by step solution:
Consider a container which is divided into two chambers. The volume of the first chamber \[{v_1} = 4.5l\] and volume of the second chamber \[{v_2} = 5.5l\]. The first chamber contains number of moles of \[{n_1} = 3mol\] at a pressure of \[{P_1} = 2.0atm\] and the second chamber contains number of moles of \[{n_2} = 4\,mol\] at a pressure of \[{P_2} = 3.0\,atm\].

When the partition is removed the mixture attains equilibrium, then diffusion takes place. They have given the common equilibrium pressure existing in the mixture is \[x \times {10^{ - 1}}atm\]. We need to find the value of x.
Then, \[V = \left( {{V_1} + {V_2}} \right)lit\] and \[n = \left( {{n_1} + {n_2}} \right)mol\]
Here, the temperature in both cases is the same.
For this, we use the ideal gas equation that is,
\[PV = nRT\]
Now, apply ideal gas equation for the first case, we get,
\[{P_1}{V_1} = {n_1}RT\]
\[ \Rightarrow {n_1} = \dfrac{{{P_1}{V_1}}}{{RT}}\]

Now, apply ideal gas equation for the second case, we get,
\[{P_2}{V_2} = {n_2}RT\]
\[ \Rightarrow {n_2} = \dfrac{{{P_2}{V_2}}}{{RT}}\]
When the partition is removed the mixture attains equilibrium, then,
\[P\left( {{V_1} + {V_2}} \right) = \left( {{n_1} + {n_2}} \right)RT\]
Now, substitute the value of \[{n_1}\]and \[{n_2}\]we get,
\[P\left( {{V_1} + {V_2}} \right) = \left( {\dfrac{{{P_1}{V_1}}}{{RT}} + \dfrac{{{P_2}{V_2}}}{{RT}}} \right)RT\]
\[\Rightarrow P\left( {{V_1} + {V_2}} \right) = \left( {\dfrac{{{P_1}{V_1} + {P_2}{V_2}}}{{RT}}} \right)RT\]
\[\Rightarrow P = \left( {\dfrac{{{P_1}{V_1} + {P_2}{V_2}}}{{{V_1} + {V_2}}}} \right)\]

Substitute the values of pressure and volume we get,
\[P = \left( {\dfrac{{2 \times 4.5 + 3 \times 5.5}}{{4.5 + 5.5}}} \right)\]
\[\Rightarrow P = \left( {\dfrac{{9 + 16.5}}{{10}}} \right)\]
\[\Rightarrow P = 2.55\]
\[ \Rightarrow P = 25.5 \times {10^{ - 1}}atm\] and the common equilibrium pressure existing in the mixture is \[x \times {10^{ - 1}}atm\]
Comparing these two we get,
\[\therefore x = 25.5\]

Therefore, the value of x is 25.5.

Note:Basically the term ideal gas refers to a hypothetical gas that is composed of molecules which follow a few rules, that is the ideal gas molecules do not attract or repel each other. The interaction between ideal gas molecules would be an elastic collision upon impact with each other or an elastic collision with the walls of the container.