Sankalp NEET Full Test-3 Question-29 Solution

Question: 29. A container has two chambers of volumes V_{1}=2 litres and V_{2}=3 litres separated by a partition made of a thermal insulator. The chambers contains n_{1}=5 and n_{2}=4 moles of ideal gas at pressures p_{1}=1 \mathrm{~atm} and p_{2}=2 \mathrm{~atm}, respectively. When the partition is removed, the mixture attains an equilibrium pressure of:

(1) 1.3 atm

(2) 1.6 atm

(3) 1.4 atm

(4) 1.8 atm

Answer: Option (2)

Explanation:

For an ideal gas, pV = nRT.

Initially, the two chambers are separated by an insulating partition,

so each side has its own temperature.

For chamber 1:

p_{1} V_{1} = n_{1} R T_{1}

So, T_{1} = \frac{p_{1} V_{1}}{n_{1} R}

For chamber 2:

p_{2} V_{2} = n_{2} R T_{2}

So, T_{2} = \frac{p_{2} V_{2}}{n_{2} R}

We do not need the actual value of R, only the ratios. Using the given data:

T_{1} \propto \frac{p_{1} V_{1}}{n_{1}} = \frac{1 \times 2}{5} = \frac{2}{5} T_{2} \propto \frac{p_{2} V_{2}}{n_{2}} = \frac{2 \times 3}{4} = \frac{6}{4} = \frac{3}{2}

When the partition is removed, the container is thermally insulated from outside,

so total internal energy of the gas mixture is conserved.

For an ideal gas, internal energy U is proportional to

nT (since U = n C_{V} T and C_{V} is same for both parts).

Thus, energy conservation gives:

n_{1} T_{1} + n_{2} T_{2} = (n_{1} + n_{2}) T_{f}

where T_{f} is the final equilibrium temperature.

Substitute values using the proportional temperatures:

n_{1} T_{1} = 5 \times \frac{2}{5} = 2 n_{2} T_{2} = 4 \times \frac{3}{2} = 6

So, n_{1} T_{1} + n_{2} T_{2} = 2 + 6 = 8

Total moles: n_{1} + n_{2} = 5 + 4 = 9

Therefore,

T_{f} = \frac{8}{9} (in the same proportional units)

Now the final pressure p_{f} after removing the partition is given by the ideal gas law for the whole container:

p_{f} V_{\text{total}} = (n_{1} + n_{2}) R T_{f}

Total volume: V_{\text{total}} = V_{1} + V_{2} = 2 + 3 = 5

So,

p_{f} = \frac{(n_{1} + n_{2}) R T_{f}}{V_{\text{total}}} = \frac{9 R \times \frac{8}{9}}{5} = \frac{8 R}{5}

To compare with the given pressures in atm,

note that our proportional calculation effectively uses

the same R and consistent units, so:

p_{f} = \frac{8}{5} = 1.6 atm

Hence, the equilibrium pressure of the gas mixture is 1.6 atm,

which corresponds to Option (2).