Question

After the completion of a gas forming reaction, the column of water remaining in the collection flask was measured to me 55mm high, the vapor pressure of water is 0.0313atm at 25oC, and the atmospheric pressure that day was measured as 0.950 atm, what is the partial pressure of the gas produced by the reaction? 1mmHg=13.595mm H2O and 1atm=760mmHg

Answer 1

The partial pressure of the gas generated in the reaction have to get.

The partial pressure of the gas generated in the reaction is 0.025 atm.      

As per the Barometric distribution law: Pressure (P) = height (h) × density (ρ) × gravitational acceleration (g).

Here we have to get the final pressure (P).

The change of height (h) =  55 mm of water or, $\frac{55}{13.595}$ = 4.045 mm of Hg [As 1 mm of Hg is equivalent to 13.595 mm of water]

Or, 0.404 cm

The gravitational acceleration (g) = 980 cm/sec²

The density of Hg (ρ) = 13.6 g/cm³

On plugging the values: P = 0.404 cm × 13.6 g/cm³ × 980 cm/sec²

P = 5384.512 dyne/cm²

Now, we know 1 dyne/cm² = 9.869×10⁻⁷ atm pressure

So, 5390 dyne/cm² = 5384.512×9.869×10⁻⁷ = 5.313×10⁻³ atm pressure.

The total vapor pressure of water is 0.0313 atm.

As per Dalton law, The total pressure = partial pressure of the water + partial pressure of the gas produced.

0.0313 = 5.313×10⁻³ + partial pressure of the gas.

Partial pressure of the gas = 0.025 atm.

Thus the partial pressure of the gas generated in the reaction is 0.025 atm.