Question

Q&A A tank contains 12.1 g of chlorine gas (Cl2) at a temperature of 84 °C and an absolute pressure of 5.40 × 105 Pa. The mass per mole of Cl2 is 70.9 g/mol. (a) Determine the volume of the tank. (b) Later, the temperature of the tank has dropped to 29 °C and, due to a leak, the pressure has dropped to 3.60 × 105 Pa. How many grams of chlorine gas have leaked out of the tank?

Answer 1

(a)
Assume ideal gas behavior:
P∙V = n∙R∙T

The number of moles in the tank equals mass divided by molar mass
n = m/M

Hence,
P∙V = (m/M)∙R∙T

So the volume of the tank is
V = (m/M)∙R∙T / (P)
= (12.1 g / 70.9 g∙mol⁻¹) ∙ 8.3145 Pa∙m³∙K⁻¹∙mol⁻¹ ∙ (84 + 273) K / 5.4×10⁵
= 9.38×10⁻⁴ m³

(b)
From ideal gas law follows
m∙T/P = V∙M/R = constant
throughout the leaking process because the volume of the tank remains unchanged.
So the state before and after leaking (with prime) are related as:
m∙T/P = m'∙T'/P'

Hence,
m' = m∙(T/T')∙(P'/P)
= 12.1 g ∙ ( (84 + 273)/(29 + 273) ) ∙ ( 3.6×10⁵./ 5.4×10⁵ )
= 9.54 g

The mass of chlorine which has leaked is:
∆m = m - m' = 12.1 - 9.54 g = 2.56 g