What pressure IN "kilopascals" is exerted by the DRY gas if it is has been collected over water at 75 ""^@C, and the total pressure is 3.41*atm?

Apr 27, 2017

You need to quote the $\text{saturated vapour pressure}$ for an accurate answer.

Explanation:

The pressure exerted by gases ABOVE an aqueous phase is equal to the SUM of the pressure of the individual gases, and the so-called $\text{saturated vapour pressure}$ that the water molecules exert. That this must be considered in the problem is indicated by the specific wording; they specify a $\text{dry gas}$.

At $75$ ""^@C, $\text{= SVP = 0.3806"*"atm}$. These data are taken from this site. What is $\text{SVP}$ at $100$ ""^@C?

And so ${P}_{\text{measured"=P_"gas"+P_"SVP}}$, and............

${P}_{\text{Gas"=P_"measured"-P_"SVP}} = \left(3.41 - 0.3806\right) \cdot a t m = 3.029 \cdot a t m$.

And so, since $1 \cdot a t m \equiv 101.3 \cdot k P a$, then...........

${P}_{\text{gas}} = 3.029 \cdot a t m \times 101.3 \cdot k P a \cdot a t {m}^{-} 1 \cong 303 \cdot k P a$.