A certain gas cylinder A has a volume of 4

We can solve this problem by using the ideal gas law. First we must convert the pressures to Pascals:

760 mm Hg corresponds to 1 atmosphere and:

1 atmosphere = 101325 Pa

Converting the pressures to pascal gives:

P_{O2} = 123.59*10^3 Pa

P_{Ne} = 119.72*10^3 Pa

P_{mixture} = 120.92*10^3 Pa

The ideal gas law states:

PV = N k T

Here N is the number of molecules in the gas and k is Boltzmann's constant:

k = 1.38065*10^(-23)J/K

T is the absolute temperature.

We know the pressure, volume and temperature of the O2 in gas cylinder A before mixing. We can use this to calculate the number of O2 molecules:

N_{O2} = P_{O2}V_{O2}/(kT) = 123.59*10^3 Pa* 4.78 L/[k * (25 + 273.15)K] =

123.59*10^3 Pa* 4.78*10^(-3) m^3/[k * (25 + 273.15)K] =

1.4351*10^23

The gas law gives the pressure in gas cylinder B before mixing as:

P_{Ne} = N_{Ne} k T/V_{Ne} (1)

and it gives the pressure after mixing as:

P_{mixture} = [N_{O2} + N_{Ne}]kT/[V_{O2} + V_{Ne}] (2)

P_{Ne} = 119.72*10^3 Pa and Eq. (1) implies that:

N_{Ne}/V_{Ne }= 2.908*10^(25)m^(-3) (3)

Note that the temperature stays the same after mixing because the temperatures in the gas cylinders were equal before mixing.

We can rewrite Eq. (2) as:

P_{mixture} = [N_{O2} + N_{Ne}]kT/[V_{O2} + V_{Ne}] =

[N_{O2}/V_{Ne} + N_{Ne}/V_{Ne}]kT/[V_{O2}/V_{Ne} + 1] ----->

[V_{O2}/V_{Ne} + 1] P_{mixture}/(kT) = N_{O2}/V_{Ne} + N_{Ne}/V_{Ne} ---->

[V_{O2}P_{mixture}/(kT) - N_{O2}]/V_{Ne} = N_{Ne}/V_{Ne} - P_{mixture}/(kT) -->

V_{Ne} = [V_{O2}P_{mixture}/(kT) - N_{O2}]/[N_{Ne}/V_{Ne} - P_{mixture}/(kT) ]

Inserting in here the figures:

V_{O2} = 4.78*10^(-3) m^3

P_{mixture} = 120.92*10^3 Pa

N_{O2} = 1.4351*10^23

and the result we found in Eq. (3):
N_{Ne}/V_{Ne} = 2.908*10^(25)m^(-3)

Gives:

V_{Ne} = 1.0494*10^(-2)m^3 = 10.5 L