Question

9. A rigid tank contains 65.5 g of chlorine gas (Cl2) at a temperature of 73 °C and an absolute pressure of 5.60 × 10 5 Pa. Later, the temperature of the tank has dropped to 40 °C and, due to a leak, the pressure has dropped to 3.50 × 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

given tank contains chlorine gas having,

m = 65.5 g

T = 73^oc = 346 K

P = 5.6 x 10⁵ Pa

M = 70.9 g/mol.

Now, to find the leakage in chlorine, we need to first find the volume using ideal gas equation-

1 9.9. Pressure (P) = 5.6*163 Pa Temp. = 973+73 : 346 * mc chlorine gas) = 65.5g. m (molecular, wt). TO-9 g/mol. n= ñ 65.5 70.9 = 0.923 moles. gaus ideal equation - By PV = nRT V MRT P - 0.923 X 8.314 X 346 5.68105 - 474.13 10 5 tank Hence, the volume of the 474.13 x105 m3

Now, the temperature changes to 40^oC and pressure to 3.50 x 10⁵ Pa.

later, Due to leakage, Temperature : 40°C =(273 +40 DK = 313k: P = 3.50 x 105 pa Using, ideal gas law, PV = nRT n - PV RT - 3.50x10*x 474.13*10% 8.314 X 313 - 0-637 Thus, mass of cla is - -0.637 X 70.9 @ 45.21 Therefore, amount of chlorine is, 65.5 - 45.21 - 120.29 grams leaked out

SO, amount of chlorine leaked is 20.29 g.