Question

Four gases were combined in a gas cylinder with these partial pressures: 3.5 atm N2, 2.8 atm O2, 0.25 atm Ar, and 0.15 atm He.

Answer 1

Answer:

This can be solved using Dalton's Law of Partial pressures. This law states that the total pressure exerted by a gas mixture is equal to the sum of the partial pressure of each gas in the mixture as if it exist alone in a container. In order to solve, we need the partial pressures of the gases given. Calculations are as follows:

Explanation:

P = 3.00 atm + 2.80 atm + 0.25 atm + 0.15 atm

P = 6.8 atm

3.5 atm = x (6.8 atm)

x = 0.51

Answer 2

Answer:

A. 6.7 atm

B. 0.522

C. 0.418

D. 0.037

Explanation:

Dalton's law of partial pressure states that in a micture of non-reacting gases, the total pressure exerted by the mixture os the sum of their individual partial pressure

Mathematically, this can be written as:

PT = P1 + P2 + P3 + .....+ Pn

Where PT is the total partial pressure

P1, P2, P3,...,Pn is the individual partial pressure of the gases that form the mixture.

A. The gases in this mixture is O2, Ar, N2 and He

O2 = 2.8 atm

Ar = 0.25 atm

N2 = 3.5 atm

He = 0.15 atm

PT = PO2 + PAr + PN2 + PHe

= (2.8 + 0.25 + 3.5 + 0.15) atm

=6.7 atm

B. Mole fraction of N2

P1 = x1*PT

PT = x1*PT + x2*PT + x3*PT +...+xnPn

Mole fraction, xn = Pn/PT

xN2 = 3.5/6.7

= 0.522

C. Mole fraction of O2

xO2 = 2.8/6.7

= 0.418

D. Mole fraction of Ar

xAr = 0.25/6.7

= 0.037