
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.
Answer
162.6k+ views
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.
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.
Recently Updated Pages
Fluid Pressure - Important Concepts and Tips for JEE

JEE Main 2023 (February 1st Shift 2) Physics Question Paper with Answer Key

Impulse Momentum Theorem Important Concepts and Tips for JEE

Graphical Methods of Vector Addition - Important Concepts for JEE

JEE Main 2022 (July 29th Shift 1) Chemistry Question Paper with Answer Key

JEE Main 2023 (February 1st Shift 1) Physics Question Paper with Answer Key

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Charging and Discharging of Capacitor

Other Pages
NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

Motion in a Straight Line Class 11 Notes: CBSE Physics Chapter 2

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line

Important Questions for CBSE Class 11 Physics Chapter 1 - Units and Measurement
