Question

if 4.03 g of Ar are added to 3.01 atm of He in a 2.00 L cylinder at 27.0 C, what is the total pressure of the resulting gaseous mixture?

Answer 1

Calculation of pressure exerted by Ar :

We know that PV = nRT

Where

T = Temperature = 27 ^oC = 27+273 = 300 K

P = pressure = ?

n = No . of moles = mass/molar mass = 4.03 g / 40 (g/mol)

     = 0.101 mol

R = gas constant = 0.0821 L atm / mol - K

V= Volume of the cylinder = 2.00 L

Plug the values we get P = (nRT)/V

                                    = (0.101 x0.0821x300)/2.00

                                   = 1.24 atm

Therefore the pressure exerted by Ar = 1.24 atm

Given the pressure exerted by He = 3.01 atm

Therefore the total pressure exerted by two gases is 1.24+3.01 = 4.25 atm