Question

. A rigid tank contains 65.5 g of chlorine gas (Cl2) at a temperature of 73 °C and an absolute pressure of 6.00 × 10 5 6.00 × 10 5 Pa. Later, the temperature of the tank has dropped to 34 °C and, due to a leak, the pressure has dropped to 3.70 × 10 5 3.70 × 10 5 Pa. How many grams of chlorine gas have leaked out of the tank? (The mass per mole of Cl2 is 70.9 g/mol.)

Answer 1

Answer:

Chlorine gas leaked = 20 g        

Explanation:

Given: Molar mass of chlorine gas: m = 70.9 g/mol

Initial Mass of chlorine gas: w₁ = 65.5 g, Initial Absolute pressure: P₁ = 6 × 10⁵ Pa, Initial temperature: T₁ = 73 °C = 73 + 273 = 346 K      (∵ 0°C = 273 K)

Final Absolute pressure: P₂ = 3.70 × 10⁵ Pa, Final temperature: T₂ = 34 °C = 34 + 273 = 307 K

Volume is constant

Final mass of chlorine gas: w₂ = ? g

Chlorine gas have leaked = w₁ - w₂ = ? g

Initial number of moles of chlorine gas: $n_{1}= \frac{w_{1}}{m_{1}} =\frac{65.5 g}{70.9 g/mol} = 0.924 mole$

According to the Ideal gas law: P.V = n.R.T

∴ at constant volume,

$\frac{n_{1}\times T_{1}}{P_{1}} = \frac{n_{2}\times T_{2}}{P_{2}}$

$n_{2} = \frac{n_{1}\times T_{1} \times P_{2}}{P_{1} \times T_{2}}$

$n_{2} = \frac{(0.924)\times (346 K) \times (3.70 \times 10^{5} Pa)}{(6 \times 10^{5} Pa) \times (307 K)}$

Final number of moles of chlorine gas: $n_{2} = 0.642 = \frac{w_{2}}{m}$

⇒ Final mass of chlorine: $w_{2} = 0.642 mol \times m = (0.642 mol) \times (70.9 g/mol) = 45.5 g$

Therefore, Chlorine gas leaked = w₁ - w₂ = 65.5 g - 45.5 g = 20 g