Exercise 3 | Properties of Gases

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 M_C + 10 x M_H = 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 = $\frac{6.58 \times {10}^5 \times 35.0 \times {10}^{-3} \times 58.1}{8.314 \times 298}$ = 5.40 x 10² g of butane