A gas mixture contains each of the following gases at the indicated partial pressures: N2, 211 torr ; O2, 139 torr ; and He, 126 torr.?
A gas mixture contains each of the following gases at the indicated partial pressures: N2, 211 torr ; O2, 139 torr ; and He, 126 torr . What mass of each gas is present in a 1.50 −L sample of this mixture at 25.0 ∘C?
- hcbiochemLv 71 month agoBest Answer
For each gas, just use the ideal gas law to calculate moles and then masses of each one.
PV = n RT
(211/760 atm) (1.50 L) = n (0.08206 Latm/mol K) (298 K)
n = 0.0170 moles
Mass N2 = 0.0170 mol X 28.0 g/mol = 0.477 grams N2
Repeat these calculations for each gas and you're there.
- pisgahchemistLv 71 month ago
Gas mixture ....
This question is asking you to compute the mass of a quantity of gas. The fact that there are several gases will not affect the calculation for each gas.
PV = nRT ............... ideal gas equation
PV = mRT / M ........ n = m / M .... m=mass, M=molar mass
m = PVM / RT ........ solve for m
m = 211 torr x 1.50L x 28.0g/mol / 62.36 Ltorr/molK / 298K
m = 0.477g N2
Since you have several more calculations to do and V, R and T are the same, collapse those into a constant: k=8.07x10^-5, and store that in a memory in your calculator. Then simply recall it for each calculation.
m(O2) = 139 torr x 32.0g/mol x k = 0.351g O2
m(He) = 126 torr x 4.0 g/mol x k = 0.0407g He
One other thing. You don't have to convert the pressure to atmospheres. You can use the value of R which matches the pressure units in the question.
R = 0.08206 Latm/molK ........ atm
R = 8.314 LkPa/molK ............ kPa
R = 0.08314 Lbar/molK ......... bar
R = 62.36 Ltorr/molK ............. torr (aka mm Hg)