Question

30m^3 of moist air at a total presssure of 101.325 kPa and 303K contains water vapour in such a proportion that its partial pressure is 2.933 kPa
Keeping the total pressure constant, temperature is reduced to 288K, during which process, some water vapour condenses out. After cooling, partial pressure of water is found to be 1.693kPa. Calculate the volume of air at 288K and weight of water condensed.

Answer 1

Let us write down the state of the gas system before and after the cooling in terms of the ideal gas equation:

$$P_{0}V_{1}=(n_{0}+n_{1})RT_{1}$$

$$P_{0}V_{2}=(n_{0}+n_{2})RT_{2}$$

, where P₀ - total pressure of the wet gas, V₁ and V₂ - the volumes of the gas before and after cooling, T₁ and T₂ - temperatures of the gas before and after cooling, n₀ - amount in moles of the pure gas without water vapor, n₁ and n₂ - the amounts of water vapors before and after cooling (condensation).

Let us now state the ideal gas equations for the water vapors before and after cooling (condensation):

$$P_{part.1}V_{1}=n_{1}RT_{1}$$

$$P_{part.2}V_{2}=n_{2}RT_{2}$$

Further, we will mathematically reorganized both groups of equation in a manner that n₀, n₁, n₂ and R will disappear:

1) Extract n₁ and n₂ from the equations with the partial water pressure and put them into the first pair of equations

$$P_{0}V_{1}=(n_{0}+\frac{P_{part.1}V_{1}}{RT_{1}})RT_{1}$$

$$P_{0}V_{2}=(n_{0}+\frac{P_{part.2}V_{2}}{RT_{2}})RT_{2}$$

Now, open the brackets and separate n₀ at the right sight of each equation

$$P_{0}V_{1}-P_{part.1}V_{1}=n_{0}RT_{1}$$

$$P_{0}V_{2}-P_{part.2}V_{2}=n_{0}RT_{2}$$

Its the right time to divide the first equation with the second one

$$\frac{V_{1}(P_{0}-P_{part.1})}{V_{2}(P_{0}-P_{part.2})}=\frac{T_{1}}{T_{2}}$$

Finally, we can extract and calculate V₂ value from the prepared relation

$$V_{2}=\frac{T_{2}V_{1}(P_{0}-P_{part.1})}{T_{1}(P_{0}-P_{part.2})}=\frac{288K 30m^3(101325 Pa-2933 Pa)}{303K(101325 Pa-1693 Pa)}=28 m^3$$

2) For the second part of the question we will use the ideal gas equations as for the water vapor partial pressures before and after the cooling

$$P_{part.1}V_{1}=\frac{m_{water.1}}{M_{water}}RT_{1}$$

$$P_{part.2}V_{2}=\frac{m_{water.2}}{M_{water}}RT_{2}$$

, where m_water.1 and m_water.2 - masses of water vapors before and after condensation, M_water - molar weight of water

Now we will extract water vapor masses from both equations and found their difference, which is in fact the mass of condensed water

$$m_{water.1}=\frac{P_{part.1}V_{1}M_{water}}{RT_{1}}$$

$$m_{water.2}=\frac{P_{part.2}V_{2}M_{water}}{RT_{2}}$$

$$m_{cond.}=m_{water.1}-m_{water.2}=\frac{M_{water}}{R}(\frac{P_{part.1}V_{1}}{T_{1}}-\frac{P_{part.2}V_{2}}{T_{2}})$$

Thus, the mass of condensed water is equal to

$$m_{cond.}=\frac{18*10^{-3}kg/mol}{8.31J/(mol*K)}(\frac{2933Pa 30m^3}{303K}-\frac{1693Pa 28m^3}{288K})=0.273kg$$

The volume at 288 K would be 28 m³ and the mass of the condensed water is equal to 273 g.