# Question #5cddc

Dec 7, 2014

The answer is $693.5 m m H g$.

The total pressure reqistered for the collected gas represents the sum of the partial pressures of water(its vapor pressure ) and hydrogen gas. We know that

${P}_{T O T A L} = {P}_{{H}_{2}} + {P}_{w a t e r}$

Usually, water's vapor pressure at various temperatures is given, but you could try and calculate it using the given temperature by

$P = \exp \left(20.386 - \frac{5132}{T}\right) m m H g$, where temperature is in K and pressure in mmHg. Using this approximation formula (wikipedia link: http://en.wikipedia.org/wiki/Vapour_pressure_of_water) we get

${P}_{w a t e r} = \exp \left(20.386 - \frac{5132}{273.15 + 30}\right) = 31.7 m m H g$

I've looked up the actual value of the vapor pressure at that temperature and it's listed at

${P}_{w a t e r} = 31.5 m m H g$, which is pretty close to the value determined using the approximation formula.

Let's use the $31.5 m m H g$ vapor pressure. This will result in

${P}_{{H}_{2}} = {P}_{T O T A L} - {P}_{w a t e r} = 725 - 31.5 = 693.5 m m H g$

Water's vapor pressure values:

http://intro.chem.okstate.edu/1515sp01/database/vpwater.html