Question

A gas mixture with a total pressure of 750 mmHg contains each of the following gases at the indicated partial pressures: CO2 , 124 mmHg ; Ar , 218 mmHg ; and O2, 197 mmHg . The mixture also contains helium gas. What is the partial pressure of the helium gas?
What mass of helium gas is present in a 13.0-L sample of this mixture at 282 K ?

Answer 1

Answer: a) 211 mm Hg

b) 0.629 grams

Explanation:

According to Dalton's law, the total pressure is the sum of individual pressures.

$p_{total}=p_1+p_2+p_3+p_4$

$p_{total}$ = total pressure = 750 mmHg

$p_{CO_2}$ = 124 mm Hg

$p_{Ar}$ = 218 mm Hg

$p_{O_2}$ = 197 mm Hg

$p_{He}$ = ?

$750 mmHg=124 mm Hg+218 mm Hg+197 mm Hg+p_{He}$

$p_{He}=211mmHg$

Thus  the partial pressure of the helium gas is 211 mmHg.

b) According to the ideal gas equation:

$PV=nRT$

P = Pressure of the gas = 211 mmHg = 0.28 atm   (760mmHg=1atm)

V= Volume of the gas = 13.0 L

T= Temperature of the gas =  282 K  

R= Gas constant = 0.0821 atmL/K mol

n= moles of gas= ?

$n=\frac{PV}{RT}=\frac{0.28\times 13.0}0.0821\times 282}=0.157moles$

Mass of helium= $moles\times {\text {molar mass}}=0.157\times 4=0.629g$

Thus mass of helium gas present in a 13.0-L sample of this mixture at 282 K is 0.629 grams