Question

9. Liquid helium at 4.2 K has a density of 0.147 g/mL. Suppose that a 2.50-L metal bottle that contains air at 125K and 1.0 atm pressure is sealed off. If we inject 130.0 mL of liquid helium and allow the entire system to warm to room temperature (25°C), what is the pressure inside the bottle?

Answer 1

Answer : The pressure inside the bottle is 49.2 atm

Explanation :

First we have to calculate the mass of helium.

`\text{Mass of helium}=\text{Density of helium}* \text{Volume of helium}=0.147g/mL* 130.0mL=19.11g`

Now we have to calculate the moles of helium.

`\text{Moles of helium}=\frac{\text{Mass of helium}}{\text{Molar mass of helium}}=(19.11g)/(4.00g/mol)=4.78mole`

Now we have to calculate the moles of air in the container.

Using ideal gas equation :

PV = nRT

where,

P = Pressure of air = 1.0 atm

V = Volume of air = 2.50 L

n = number of moles of air = ?

R = Gas constant =
`0.0821L.atm/mol.K`

T = Temperature of air = 125 K

Putting values in above equation, we get:

`1.0atm* 2.50L=n* 0.0821L.atm/mol.K* 125K\\\\n=0.24`

Now we have to calculate the pressure of individual components at
`25^oC(298K)`.

Pressure of helium:

`P_(He)=(nRT)/(V)`

`P_(He)=((4.78mol)* (0.0821L.atm/mol.K)* (298K))/(2.50L)=46.8atm`

Pressure of air :

`P_(air)=(nRT)/(V)`

`P_(air)=((0.24mol)* (0.0821L.atm/mol.K)* (298K))/(2.50L)=2.4atm`

The overall pressure =
`P_(He)+P_(air)=46.8+2.4=49.2atm`

Therefore, the pressure inside the bottle is 49.2 atm