Question

850.0 gra of flourine gas F2 is removed from a 200.0 L-tank ay 40.0 C° and 32.0 atm without changing the volume of temperature .Calculate the new pressure in the tank

Answer 1

Given the volume of the tank, V = 200.0 L

Temperature, T = 40.0 DegC = 40+273 = 313 K

Initial pressure, Pi = 32.0 atm

Initial moles of F2 gas, n1 can be calculated from ideal gas equation.

PV = n1RT

=> n1 = PV / RT

=> n1 = 32.0 atm x 200.0 L / (0.0821 L.atm.mol-1.K-1 x 313 K) = 249.05 mol

moles of F2 gas removed = mass / molar mass = 850.0 g / 38.0 g/mol = 21.8 mol

Hence final moles of F2 gas, n2 = 249.05 mol - 21.8 mol = 227.25 mol

Since V and T are constant,

P / n = constant

=> P1 / n1 = P2 / n2

=> 32.0 atm / 249.05 mol = P2 / 227.25 mol

=> P2 = 29.2 atm (answer)