Question

A 2.57-L flexible flask at 12°C contains a mixture of N2, He, and Ne at partial pressures of 0.297 atm for N2, 0.157 atm for He, and 0.455 atm for Ne. (a) Calculate the total pressure of the mixture. (b) Calculate the volume in liters at STP occupied by He and Ne if the N2 is removed selectively.

Answer 1

a)

Total pressure = sum of all partial pressures

= 0.297 atm + 0.157 atm + 0.455 atm

= 0.909 atm

Answer: 0.909 atm

b)

for He:

Given:

Pi = 0.157 atm

Pf = 1 atm

Vi = 2.57 L

Ti = 12.0 oC

= (12.0+273) K

= 285 K

Tf = 273.0 K

use:

(Pi*Vi)/(Ti) = (Pf*Vf)/(Tf)

(0.157 atm*2.57 L)/(285.0 K) = (1 atm*Vf)/(273.0 K)

Vf = 0.3865 L

Answer: 0.386 L

For Ne:

Given:

Pi = 0.455 atm

Pf = 1 atm

Vi = 2.57 L

Ti = 12.0 oC

= (12.0+273) K

= 285 K

Tf = 273.0 K

use:

(Pi*Vi)/(Ti) = (Pf*Vf)/(Tf)

(0.455 atm*2.57 L)/(285.0 K) = (1 atm*Vf)/(273.0 K)

Vf = 1.12 L

Answer: 1.12 L

for N2:

Given:

Pi = 0.297 atm

Pf = 1 atm

Vi = 2.57 L

Ti = 12.0 oC

= (12.0+273) K

= 285 K

Tf = 273.0 K

use:

(Pi*Vi)/(Ti) = (Pf*Vf)/(Tf)

(0.297 atm*2.57 L)/(285.0 K) = (1 atm*Vf)/(273.0 K)

Vf = 0.7312 L

Answer: 0.731 L