Question

A gas mixture with a total pressure of 755 mmHg contains each of the following gases at the indicated partial pressures: 126 mmHg CO2, 205 mmHg Ar, and 192 mmHg O2. The mixture also contains helium gas. What mass of helium gas is present in a 10.4-L sample of this mixture at 265 K ?.

Answer 1

Solution

Given data volume = 10.4 L

Total pressure = 755 mmHg

Temperature = 265 K

Partial pressure of CO2 = 126 mmHg

Partial pressure of Ar = 205 mmHg

Partial pressure of O2 = 192 mmHg

Mass of He gas = ?

Let first calculate the partial pressure of the He gas

Partial pressure of He = total pressure – ( pCO2 +pAr + pO2)

Partial pressure of He = 755 mmHg – ( 126 mmHg + 205 mmHg + 192 mmHg)

Partial pressure of He = 228 mmHg

Now lets convert this pressure of He from mmHg to atm

( 228 mmHg * 1 atm ) / 760 mmHg = 0.305 atm

Now lets calculate the moles of the He gas using the ideal gas equation

PV= nRT

Where, P= pressure , V= volume , n= moles of gas , R= gas constant (0.08206 L atm per mol .K)

T=kelvin temperature

Now lets calculate the moles

0.305 atm * 10.4 L = n * 0.08206 L atm per mol .K * 265 K

(0.305 atm * 10.4 L) / (0.08206 L atm per mol .K * 265 K) = n

0.1435 mol = n

Therefore moles of He gas present are 0.1435 moles

Now lets convert moles of He gas to its mass

Mass =moles * molar mass

Mass of He gas = 0.1435 mol * 4.0026 g per mol

                           = 0.574 g He

Therefore mass of He gas present = 0.574 g