Question

Liquid helium at 4.2 K has a density of 0.147 g/mL. Suppose that a 3.00-L metal bottle that contains air at 99K and 3.0 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

Calculation of pressure of air :

Since the volume of the container is same that means the volume is same, so Pressure is proportional to Temperature.

P / T = P' / T'
Where P = initial pressure = 3.0 atm

            T = initial temperature = 99 K

            P' = final pressure = ?

            T' = final temperature = 25 ^oC = 25+273 = 298 K

Plug the values we get P' = PT' / T

                                         = ( 3.0 x 298 ) / 99

                                         = 9.03 atm

This is the pressure exerted by air = 9.03 atm.

Calculation of pressure of Helium gas :

Mass of He gas , m = volume x density

                                = 130.mL x 0.147 g/mL

                                = 19.11 g

Number of moles of He , n = mass / molar mass

                                           = 19.11 g / (4 g/mol)

                                           = 4.78 mol

Temperature of the system , T = 25 oC = 25+273 = 298 K

Volume of the bottle = 3.00 L

So the pressure exerted by He gas , P = nRT / V

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

Plug the above values we get P = ( 4.78 mol x 0.0821 Latm /(mol-K) x298 K ) / 3.00 L

                                                    = 38.9 atm

This is the pressure exerted by He gas = 38.9 atm

Therefore the total pressure = pressure exerted by air + Pressure exerted by He gas

                                             = 9.03 + 38.9 atm

                                             = 47.93 atm