The pressure of a moist gas at $$27°C$$ is $$4 \text{ atm}$$. The volume of the container is doubled at the same temperature. The new pressure of the moist gas is ______ $$\times 10^{-1}$$ atm. (Nearest integer)
(Given: The vapour pressure of water at $$27°C$$ is $$0.4 \text{ atm}$$)

We need to find the new pressure of the moist gas when the volume is doubled.

Initially, the total pressure of the moist gas is 4 atm at 27°C, and the vapour pressure of water at this temperature is 0.4 atm.

Subtracting the vapour pressure from the total gives the pressure of the dry gas as Total pressure - Vapour pressure of water = $$4 - 0.4 = 3.6$$ atm.

Applying Boyle’s law to the dry gas, we use $$P_1V_1 = P_2V_2$$, where $$3.6 \times V = P_2 \times 2V$$, yielding $$P_2 = \dfrac{3.6}{2} = 1.8 \text{ atm}$$.

The vapour pressure of water depends only on temperature, not on volume (as long as liquid water is present), so at 27°C it remains 0.4 atm.

In a moist gas problem, we assume water vapour remains saturated (liquid water is present), and therefore the vapour pressure stays at 0.4 atm.

The new total pressure is the sum of the pressure of dry gas and the vapour pressure of water, giving $$1.8 + 0.4 = 2.2 \text{ atm}$$.

Expressing in the required format: $$2.2 \text{ atm} = 22 \times 10^{-1} \text{ atm}$$

Hence, the answer is 22.