Question

A rigid tank contains 61.5 g of chlorine gas (Cl₂) at a temperature of 77°C and an absolute pressure of 6.00 × 10⁵ Pa. Later, the temperature of the tank has dropped to 34°C and, due to a leak, the pressure has dropped to 3.90 × 10⁵ Pa. How many grams of chlorine gas have leaked out of the tank? (The mass per mole of Cl₂ is 70.9 g/mol.)

Answer 1

We know that PV = nRT

                     PV = (w/M)RT

Where P - Pressure , V-Volume , n= number of moles , R = gas constant , T = temperature , m = mass of gas ,

M = molar mass

As the gas is same & Volume remains constant P/wT = constant

So P/wT = P'/ w'T'

Where

P = initial pressure = 6.00x10⁵ Pa

w = initial mass of Cl₂ = 61.5 g

T = initial temperature = 77^oC = 77+273 = 350 K

P' = final pressure = 3.90x10⁵ Pa

w' = final mass of Cl₂ = ?

T' = final temperature = 34^oC = 34+273 = 307 K

Plug the values we get

P/wT = P'/ w'T'

w' = (P'Tw)/(PT')

   = (3.90x10⁵ x350x61.5) / (6.00x10⁵ x307)

   = 45.6 g

Therefore the mass of chlorine gas leaked out is = initial mass - final mass

                                                                       = 61.5 - 45.6

                                                                       = 15.9 g