Question

A volume of 10 ml of ethane gas is mixed with 40 ml of oxygen gas in an eudiometer tube at 30°C and fired. When the resulting gases are cooled to 30°C, the volume of eudiometer becomes 26 ml. What is the vapour pressure of water at 30°C? Neglect the volume occupied by liquid water. Pressure is 1 atm and constant throughout.

• A. 1 atm
• B. 29.23 mm Hg
• C. 26 mm Hg
• D. 32.55 mm Hg

Answer 1

Answer: (B)

29.23 mm Hg

Answer 2

The combustion of ethane ($C_2H_6$) with oxygen ($O_2$) is: $2 C_2H_6(g) + 7 O_2(g) \rightarrow 4 CO_2(g) + 6 H_2O(g)$ 10 ml of $C_2H_6$ requires 35 ml of $O_2$ for complete combustion, producing 20 ml of $CO_2$ and 30 ml of $H_2O$ (as vapor at 30°C). After reaction, the remaining gases are 5 ml of $O_2$ and 20 ml of $CO_2$, totaling 25 ml of dry gas. The final volume is 26 ml, which includes 25 ml of dry gas and 1 ml of water vapor at 1 atm. By Dalton's Law, the partial pressure of water vapor ($P_{H_2O}$) is its volume fraction of the total volume at the total pressure: $P_{H_2O} = \frac{1 \text{ ml}}{26 \text{ ml}} \times 1 \text{ atm} = \frac{1}{26} \text{ atm}$ Converting to mm Hg (1 atm = 760 mm Hg): $P_{H_2O} = \frac{1}{26} \times 760 \text{ mm Hg} \approx 29.23 \text{ mm Hg}$