# Exercise 3 | Properties of Gases

Calculate the mass of butane C4H10 (g) in a 35.0 L container at a pressure of 6.50 atm and a temperature of 25.0°C.

We consider that butane is an ideal gas.

Ideal-gas equation:

PV = nRT    and   n = m / M

⇒ PV = $\frac{\mathrm{mRT}}{\mathrm{M}}$

⇒ m = $\frac{\mathrm{PVM}}{\mathrm{RT}}$

P = pressure in Pa ⇒ 6.50 atm = 6.58 x 105 Pa

V = volume in m3 ⇒ 35.0 L = 35.0 dm3 = 35.0 x 10-3 m3

M = molar mass ⇒ M = 4 x MC + 10 x MH = 58.1 g.mol-1

R = ideal gas constant ⇒ R = 8.314 m3.Pa.K-1.mol-1

T = temperature in K ⇒ 25.0°C = 298 K

m =  = 5.40 x 102 g of butane