A container is divided into two chambers by a partition.

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 $2.0 \mathrm{~atm}$ and second chamber contain $4.0$ moles of gas at pressure $3.0 \mathrm{~atm}$. After the partition is removed and the mixture attains equilibrium, then, the common equilibrium pressure existing in the mixture is $\mathrm{x} \times 10^{-1}$ atm. Value of $\mathrm{x}$ is

Solution:

(25)

By energy Conservation $\frac{3}{2} n_{1} R T_{1}+\frac{3}{2} n_{2} R T_{2}=\frac{3}{2}\left(n_{1}+n_{2}\right) R T$

Using $\mathrm{PV}=\mathrm{nRT}$

$\mathrm{P}_{1} \mathrm{~V}_{1}+\mathrm{P}_{2} \mathrm{~V}_{2}=\mathrm{P}\left(\mathrm{V}_{1}+\mathrm{V}_{2}\right)$

$P=\frac{P_{1} V_{1}+P_{2} V_{2}}{V_{1}+V_{2}}=\frac{2 \times 4.5+3 \times 5.5}{4.5+5.5}$

$P=\frac{9+16.5}{10}=\frac{25.5}{10}$

$\approx 25 \times 10^{-1} \mathrm{~atm}$

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