Answer:
The volume of the dry gas at STP is = 0.4314 L
Explanation:
We are given:
Total vapor pressure = 740 mmHg
Also, considering Vapor pressure of water = 23.78 mmHg
Vapor pressure of gas = Total vapor pressure - Vapor pressure of water = (740 - 23.78) mmHg = 716.22 mmHg
We use the equation given by ideal gas which follows:
[tex]PV=nRT[/tex]
where,
P = pressure of the gas = 716.22 mmHg
V = Volume of the gas = 500 mL = 0.5 L ( 1 ml = 0.001 L )
T = Temperature of the gas = [tex]25^oC=[25+273]K=298K[/tex]
R = Gas constant = [tex]62.3637\text{ L.mmHg }mol^{-1}K^{-1}[/tex]
n = number of moles of gas = ?
Putting values in above equation, we get:
[tex]716.22mmHg\times 0.5L=n\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 298K\\\\n=\frac{716.22\times 0.5}{62.3637\times 298}=0.01926mol[/tex]
At STP, 1 mole of gas yields a volume of 22.4 L
So,
0.01926 mole of gas yields volume of [tex]0.01926\times 22.4\ L[/tex]
Hence, the volume of the dry gas at STP is = 0.4314 L
