400 ml of oxygen are collected over water at 25oC and 0.948 bar pressure. What is the volume of dry gas at STP (Aq. tension at 25o C = 0.0318 bar)?
This problem can be solved using the combined gas law:
P1V1/T1 = P2V2/T2
where P1, V1, and T1 are the pressure, volume and temperature at state one, and P2,V2, T2 are the corresponding values at state 2.
To find the pressure of the dry gas you have to subtract the partial pressure of the water vapor which is given as 0.0318 bar.
So the initial conditions are:
P1 = 0.948 - 0.0318 = 0.9162 bar
V1 = 400 mL
T1 = 25 degrees C = 298.15 K
P2 = 0.986 barr
T2 = 273.15 K
V2 = unknown
substituting the values gives you
0.9162*400/298.15 = 0.986*V2/273.15
then V2 = 365.2 mL
STP stands for Standard Temperature & Pressure
The standard temperature is 273.15 Kelvin while the standard pressure is 1 bar = 1.01325 atm
Now, As per the Ideal Gas Equation; P*V = n*R*T;
where P = pressure of the gas in atm, V = volume of the gas in litres , n = moles of the gas ; R = Universal Gas Constant ; T = Temperature of the gas in Kelvin
0 degrees Celsius = 273.15 Kelvin
Now, moles of the gas remains the same and so does the value of 'R' at STP and at 25oC and 0.948 bar
Thus, At both the conditions, (P*V)/T = constant
Thus, (P1*V1)/T1 = (P2*V2)/T2
Now, pressure of the gas at 25 degrees Celsius = Toatl pressure - Aqueous Tension = 0.948 - 0.0318 = 0.9162
or, (0.9162*1.01325*0.4)/298.15 = (1.01325*V2)/273.15
or, V2 = Volume of the gas at STP = 0.336 Litres = 336 ml.