A container is divided into two chambers by a partition. The volume of first chamber is 4 . 5   litre and second chamber is 5 . 5   litre . The first chamber contain 3 . 0 moles of gas at pressure 2 . 0   atm and second chamber contain 4 . 0 moles of gas at pressure 3 . 0   atm…

Question

A container is divided into two chambers by a partition. The volume of first chamber is 4 . 5 &#160; litre and second chamber is 5 . 5 &#160; litre . The first chamber contain 3 . 0 moles of gas at pressure 2 . 0 &#160; atm and second chamber contain 4 . 0 moles of gas at pressure 3 . 0 &#160; atm . After the partition is removed and the mixture attains equilibrium, then, the common equilibrium pressure existing in the mixture is x &#215; 10 - 1 &#160; atm . Value of x (nearest integer) is

Aicharya · Accepted Answer

Let common equilibrium pressure of mixture is $P$ atmp. then
$\frac{f}{2} P_{1} V_{1} + \frac{f}{2} P_{2} V_{2} = \frac{f}{2} P (V_{1} + V_{2})$
$\frac{f}{2} (2) (4.5) + \frac{f}{2} (3) (5.5) = \frac{f}{2} P (4.5 + 5.5)$
$\Rightarrow P = 2.55 = x \times 10^{- 1}$ atmp
So $x = 25.5 \approx 26$ (Nearest integer)

A container is divided into two chambers by a partition. The volume of first chamber is 4 . 5   litre…

Solution

A container is divided into two chambers by a partition. The volume of first chamber is 4 . 5 &#160; litre…

Solution

A container is divided into two chambers by a partition. The volume of first chamber is 4 . 5 litre…