Question

A mixture of He, Ar, and Xe has a total pressure of 2.90atm . The partial pressure of He is 0.300atm , and the partial pressure of Ar is 0.450atm . What is the partial pressure of Xe?

Express your answer to three significant figures and include the appropriate units.

A volume of 18.0 L contains a mixture of 0.250 mole N2 , 0.250 mole O2 , and an unknown quantity of He. The temperature of the mixture is 0 ?C , and the total pressure is 1.00 atm . How many grams of helium are present in the gas mixture?

Express your answer to three significant figures and include the appropriate units.

A mixture of helium, nitrogen, and oxygen has a total pressure of 759mmHg . The partial pressures of helium and nitrogen are 244mmHg and 196mmHg , respectively.

What is the partial pressure of oxygen in the mixture?

Answer 1

If the mole fraction of both N2 and O2 are 0.250 (sum = 0.500), then the rest of the mixture is helium (mole fraction He = 1.000 - 0.500 = 0.500).

P He = (mole fraction He)(P total) = (0.500)(1.00 atm) = 0.500 atm He

We can now use the ideal gas law to calculate moles of He, and convert that to grams of He.

PV = nRT

n He = (P He)(V) / RT = (0.500 atm)(18.0 L) / (0.0821 L atm / K mole)(273 K) = 0.402 moles He

0.402 moles He x (4.00 g He / 1 mole He) = 1.61 g He

total pressure = PO2 + PN2 +PHe

   759    = PO2 +196+244    

PO2            = 319 mmhg