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Dalton's Law of Partial Pressures

A container i...

Question

A container is divided into two chambers by a partition. The volume of the first chamber is $4.5 l i t r e s$ and the second chamber is $5.5 l i t r e s$ . The first chamber contains $3.0 m o l e$ of gas at pressure $2.0 a t m$ and the second chamber contain $4.0 m o l e$ of identical gas at pressure $3.0 a t m$ . After the partition is removed and the mixture attains equilibrium, then, the common equilibrium pressure existing in the mixture is $x \times 10^{- 1} a t m$ . Value of $x$ is _______.

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Solution

Step 1. Given data:
$V_{1} = 4.5 l i t r e s$ , Volume of first chamber
$V_{2} = 5.5 l i t r e s$ , Volume of second chamber
$n_{1} = 3 m o l e s$ , Number of moles of gas in first chamber
$n_{2} = 4 m o l e s$ , Number of moles of gas in second chamber
$P_{1} = 2 a t m$ , Pressure of gas in first chamber
$P_{2} = 3 a t m$ , Pressure of gas in second chamber
$V = V_{1} + V_{2}$ , Volume of the container
$n = n_{1} + n_{2}$ , Total number of moles in the container
Step 2. Applying gas equation separately,
For gas in the first chamber,
$P_{1} V_{1} = n_{1} R T n_{1} = \frac{P_{1} V_{1}}{R T}$ _______ $(1)$
For gas in the second chamber,
$P_{2} V_{2} = n_{2} R T n_{2} = \frac{P_{2} V_{2}}{R T}$ _______ $(2)$
Step 3. Removing partition from the container,
Applying gas equation when partition is removed,
$P V = n R T P (V_{1} + V_{2}) = (n_{1} + n_{2}) R T$
Putting value of equation $(1)$ and $(2)$ in the above equation,
$\begin{array}{rcl} \Rightarrow & P (V_{1} + V_{2}) = (n_{1} + n_{2}) R T \\ \Rightarrow & P (V_{1} + V_{2}) = (\frac{P_{1} V_{1}}{R T} + \frac{P_{2} V_{2}}{R T}) R T \\ \Rightarrow & P = \frac{P_{1} V_{1} + P_{2} V_{2}}{(V_{1} + V_{2})} \end{array}$
Putting all values in the above equation,
$\begin{array}{rcl} \Rightarrow & P = \frac{2 \times 4.5 + 3 \times 5.5}{(4.5 + 5.5)} \\ \Rightarrow & P = \frac{25.5}{10} \\ ∴ P & = & 25.5 \times 10^{- 1} a t m \end{array}$
The value of $x$ to the nearest integer is $x = 26$ .

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