Question

A sample of butane gas C4H10 of unknown mass is contained in a vessel of unknown volume V at 25oC and a pressure of 760 mmHg. To this vessel, 8.6787 g of neon gas is added in such a way that no butane is lost from the vessel. The final pressure in the vessel is 1920 mm Hg at the same temperature. Calculate the volume of the vessel and the mass of butane.

Answer 1

mass of butane = 16.28 g

Volume = 6.85 L

Explanation:

Given,

P total= 1920 mmHg, p₁ = 760

According to Dalton's law of partial pressure,

P total = p₁ + p₂

p₂ = 1920 - 760 = 1160 mmHg

p₂ = x₂ × P total

x₂ = 1160/1920 = 0.60

moles of neon gas = mass added/ molar mass

                               = 8.6787/20.2      

                                = 0.43

x₂ = n₂/ n₁ + n₂

0.60 = 0.43/n₁ + 0.43

n₁ = 0.28

mass of butane = n₁ × molar mass

                          = 0.28 × 58

                          = 16.28 g

Convert pressure into atm pressure,and tempreture into K

PV = nRT

V = n₁RT/P

  = 0.28 × 0.0821 × 298/1

  = 6.85 L