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. After the partition is removed and the mixture attains equilibrium, then, the common equilibrium pressure existing in the mixture is x × 10-1 atm. Value of x is-

Question

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. After the partition is removed and the mixture attains equilibrium, then, the common equilibrium pressure existing in the mixture is x × 10-1 atm. Value of x is- (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Let common equilibrium pressure of mixture is

P

atmp. then

U_{1} + U_{2} = U_{mixutre}

\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 1 0^{- 1} a t m p

So

x = 25.5 \approx 26

(Nearest integer)

Let common equilibrium pressure of mixture is P atmp. then U1​+U2​=Umixutre ​ 2f​P1​V1​+2f​P2​V2​=2f​P(V1​+V2​) 2f​(2)(4.5)+2f​(3)(5.5)=2f​P(4.5+5.5) ⇒P=2.55=x×10−1atmp So x=25.5≈26 (Nearest integer)