Two containers contain the same gas at different temperatures and pressures, as indicated...

SKETCH

The situation can be represented with a neat sketch as follows.

RESEARCH

Ideal gas law can be represented by

\(p V=n R T \ldots \ldots\) (1)

Here \(p\) is the pressure of the gas, \(V\) is the volume occupied by the gas, \(n\) is the number of moles of the gas, \(R\) is the universal gas constant, and \(T\) is the absolute temperature of the

gas.

SIMPLIFY

The number of moles of the gas in container 1 is say \(n_{1}\), and the number of moles of the gas in container 2 is say \(n_{2}\).

Then the total number of moles at equilibrium is given by

\(n_{f}=n_{1}+n_{2} \ldots \ldots\) (2)

From the ideal gas law, we have

\(n_{1}=\frac{p_{1} V_{1}}{R T_{1}}\)

\(n_{2}=\frac{p_{2} V_{2}}{R T_{2}}\)

\(n_{f}=\frac{p_{f} V_{f}}{R T_{f}}\)

On substituting the values of \(n_{1}, n_{2}\) and \(n_{f}\) in equation (2), the final pressure can be obtained as

\(\frac{p_{f} V_{f}}{R T_{f}}=\frac{p_{1} V_{1}}{R T_{1}}+\frac{p_{2} V_{2}}{R T_{2}}\)

\(p_{f}=\frac{T_{f}}{V_{f}}\left(\frac{p_{1} V_{1}}{T_{1}}+\frac{p_{2} V_{2}}{T_{2}}\right)\)

CALCULATE

Pressure in container \(1, \quad p_{1}=3.00 \times 10^{5} \mathrm{~Pa}\)

Volume in the container \(1, V_{1}=2.00 \mathrm{~L}\)

Temperature in container \(1, T_{1}=600 \mathrm{~K}\)

Pressure in container \(2, \quad p_{2}=2.00 \times 10^{5} \mathrm{~Pa}\)

Volume in the container \(2, V_{2}=1.00 \mathrm{~L}\)

Temperature in the container \(2, T_{2}=200 \mathrm{~K}\)

At equilibrium the final volume of the gas is \(V_{f}=V_{1}+V_{2}\)

\(=2.00 \mathrm{~L}+1.00 \mathrm{~L}\)

\(=3.00 \mathrm{~L}\)

Final temperature of the gas at equilibrium, \(T_{f}=300 \mathrm{~K}\)

Substituting all these values in equation (3), the final pressure of the gas at equilibrium is:

$$ \begin{aligned} p_{f} &=\left(\frac{300 \mathrm{~K}}{3.00 \mathrm{~L}}\right)\left(\frac{\left(3.00 \times 10^{5} \mathrm{~Pa}\right)(2.00 \mathrm{~L})}{600 \mathrm{~K}}+\frac{\left(2.00 \times 10^{5} \mathrm{~Pa}\right)(1.00 \mathrm{~L})}{200 \mathrm{~K}}\right) \\ &=100(1000+1000) \mathrm{Pa} \\ &=200 \times 10^{3} \mathrm{~Pa} \\ &=200 \mathrm{kPa} \end{aligned} $$