Thermal Properties of Matter

JEE Main level, numerical answer, medium

70 of 126

Question 70 of 126

A container is divided into two chambers by a partition. The volume of first chamber is 4.5litre4.5litre and second chamber is 5.5litre5.5litre. The first chamber contain 3.03.0 moles of gas at pressure 2.0atm2.0atm and second chamber contain 4.04.0 moles of gas at pressure 3.0atm3.0atm. After the partition is removed and the mixture attains equilibrium, then, the common equilibrium pressure existing in the mixture is x×10−1atmx\times {10}^{-1}atm. Value of xx (nearest integer) is

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Answer: 26

Let common equilibrium pressure of mixture is PP atmp. then
f2P1V1+f2P2V2=f2P(V1+V2)\dfrac{f}{2}{P}_{1}{V}_{1}+\dfrac{f}{2}{P}_{2}{V}_{2}=\dfrac{f}{2}P\left({V}_{1}+{V}_{2}\right)
f2(2)(4.5)+f2(3)(5.5)=f2P(4.5+5.5)\dfrac{f}{2}\left(2\right)\left(4.5\right)+\dfrac{f}{2}\left(3\right)\left(5.5\right)=\dfrac{f}{2}P\left(4.5+5.5\right)
⇒P=2.55=x×10−1\Rightarrow P=2.55=x\times {10}^{-1} atmp
So
x=25.5≈26x=25.5\approx 26 (Nearest integer)

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