Question

A 1.00 L flask is filled with 1.15 g of argon at 25 ∘C. A sample of ethane vapor is added to the same flask until the total pressure is 1.200 atm .

A) What is the partial pressure of argon, PAr, in the flask?

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

B) What is the partial pressure of ethane, Pethane, in the flask?

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

Answer 1

a)

Partial Pressure Argon = x-Argon * Ptotal

initially

mol of Argon = mass/MW = 1.15/39.948 = 0.0287 mol of Argon

Apply Ideal Gas Law,

PV = nRT

where

P = absolute pressure

V = total volume of gas

n = moles of gas

T = absolute Temperature

R = ideal gas constant

PV = nRT

n = PV/(RT) = (1.2)(1)/(0.082*298)=0.049107 mol of gases

mol of Ar --> 0.0287

x-Ar = mol of AR / Total mol = (0.0287)/0.049107 = 0.5844

P-Ar = 0.5844 * 1.2

P-Ar = 0.70128 atm --> 0.701 atm

b)

for ethane

P-ethane = Ptotal - Pargon = 1.20-0.701 = 0.499 atm