Better Students Ask More Questions.
The mole fraction of argon in a gas mixture, which contains only argon and hydrogan,...
The mole fraction of argon in a gas mixture, which contains only argon and hydrogan, is
0.650. Calculate the density of the gas mixture at a pressure of 1.50 bar and a temperature of
3.00 × 102 K. State the assumption you used in this calculation.Relative atomic masses:Ar =40,H =1
1 Answer | add yours
- Gasses behave as ideal gasses
- Gasses don't react with each other
- No internal bonds between gas molecules or all bonds are equal
Best answer as selected by question asker.
Here we can use ideal gas equation for the calculation.
`n = (PV)/(RT)`
P = Pressure
V = Volume
n = Moles
R = Universal gas constant
T = temperature
Also we know that;
`P_A = X_AP_T`
`P_A` = Partial pressure of A
`X_A` = mole fraction of A
`P_T ` = Total pressure
Assume the volume to be `1m^3` .
`1 bar = 1xx10^5pa`
`n = (0.65xx1.5xx10^5)/(8.314xx300)`
`n = 39.09mol`
For `H_2` ;
`n = (0.35xx1.5xx10^5)/(8.314xx300)`
`n = 21.05mol`
Mass of Ar `= 39.09xx40 = 1593.6`
Mass of `H_2` `= 21.05xx2 = 42.1`
Total mass `= 1593.6+42.1 = 1635.7g = 1.64kg`
Total Volume `= 1m^3`
Density of the mixture = 1.64kg/m^3
Posted by jeew-m on May 7, 2013 at 7:52 AM (Answer #1)
Join to answer this question
Join a community of thousands of dedicated teachers and students.