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#?

Question

Wyatt Adkins · Accepted Answer

You need to quote the #"saturated vapour pressure"# for an accurate answer.

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 #"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 #"dry gas"#. At #75# #""^@C#, #"= SVP = 0.3806"*"atm"#. These data are taken from this site. What is #"SVP"# at #100# #""^@C#? And so #P_"measured"=P_"gas"+P_"SVP"#, and............ #P_"Gas"=P_"measured"-P_"SVP"=(3.41-0.3806)*atm=3.029*atm#. And so, since #1*atm-=101.3*kPa#, then........... #P_"gas"=3.029*atmxx101.3*kPa*atm^-1~=303*kPa#.

Wyatt Adkins · Answer

undefined The vapor pressure of water at 75°C is approximately 28.3 kPa. To find the pressure exerted by the dry gas, we subtract the vapor pressure of water from the total pressure:

$ \text{Pressure of dry gas} = \text{Total pressure} - \text{Vapor pressure of water} $

Plugging in the values:

$ \text{Pressure of dry gas} = 3.41 \text{ atm} - 28.3 \text{ kPa} $

$ \text{Pressure of dry gas} \approx 315.7 \text{ kPa} $

So, the pressure exerted by the dry gas is approximately 315.7 kilopascals.