Question

Chapter 14, Problem 24 GO A tank contains 12.9 g of chlorine gas (Cl2) at a temperature of 83 °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 25 °C and, due to a leak, the pressure has dropped to 3.40 × 105 Pa. How many grams of chlorine gas have leaked out of the tank? (a) Number Units (b) Number Units

Answer 1

(a) Use the ideal gas expression -

P?V = n?R?T

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

Therefore -

P?V = (m/M)?R?T

So the volume of the tank is
V = (m/M)??R?T / (M?P)

put the values -

V = (12.9 g / 70.9 g?mol?¹) ? 8.3145 Pa?m³?K?¹?mol?¹ ? (83 + 273) K / 5.40×10?
= 9.97×10?? m³
= 0.997 L

(b) Again 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'

Therefore -

m' = m?(T/T')?(P'/P)
= 12.9 g ? ( (25 + 273)/(83 + 273) ) ? ( 3.40×10?./ 5.40×10? )
= 12.9 *(298 / 356)* 0.6296 = 6.80 g

Therefore, the mass of chlorine which has leaked, ?m = m - m' = 12.9 - 6.8 = 6.1 g