Answer to Question #251433 in Molecular Physics | Thermodynamics for Bebe

Question

The gaseous mixture has due point temperature of 15°C. The total pressure is 143.27 kPa. Determine the amount of water vapor present in 100-moles of the mixture. Note: Saturation pressure is at 15°C is 1.7051 kPa.

Answer 1

With the partial pressure of water and the total pressure, we can calculate the molar percentage of water:

$\chi_{_{H_2O}}=\cfrac{P_{_{H_2O}}}{P_{_{total}}}=\cfrac{ \text{1.7051 kPa} }{ \text{143.27 kPa}}=0.0119$

1.19 % of the total (100 moles) is composed of water, thus we can calculate the amount of water by using the relation with the moles of water, and then we can relate it to the amount of water:

$\chi_{_{H_2O}}=\cfrac{n_{_{H_2O}}}{n_{_{total}}} \implies n_{_{H_2O}}=\chi_{_{H_2O}}n_{_{total}} \\ m_{_{H_2O}}=\cfrac{P_{_{H_2O}}}{P_{_{total}}} \cdot n_{_{total}}M_{_{H_2O}}$

We proceed to substitute and we find

$\\ m_{_{H_2O}}= \Bigg(\cfrac{ \text{1.7051 kPa} }{ \text{143.27 kPa}} \Bigg)(100\,mol)(18\,g/mol) \\ m_{_{H_2O}}=21.4223\text{ g of water}$

In conclusion, 21.4223 g of water vapor was present in 100-moles of the mixture.

Reference

• Chang, R., & Goldsby, K. A. (2010). Chemistry. Chemistry, 10th ed.; McGraw-Hill Education: New York, NY, USA.