What pressure (in atm) would be exerted by 76 g of fluorine gas (f2) in a 1.50 liter vessel at -37oc? (a) 26 atm(b) 4.1 atm(c) 19,600 atm(d) 84(e) 8.2 atm?

Let's assume that the F ₂ gas has ideal gas behavior.  Then we can use ideal gas formula, PV = nRT Where, P is the pressure of the gas (Pa), V is the volume of the gas (m³), n is the number of moles of gas (mol), R is the universal gas constant ( 8.314 J mol ⁻¹ K⁻¹) and T is temperature in Kelvin. Moles = mass / molar mass Molar mass of F₂ = 38 g/molMass of F₂  = 76 gHence, moles of F₂ = 76 g / 38 g/mol = 2 mol P = ? V = 1.5 L = 1.5 x 10 ⁻³ m³n = 2 molR = 8.314 J mol⁻¹ K⁻¹ T = -37 °C = 236 K By substitution, P x 1.5 x 10⁻³ m³ = 2 mol x 8.314 J mol⁻¹ K⁻¹ x 236 K                         p = 2616138.67 Pa                         p = 25.8 atm = 26 atm Hence, the pressure of the gas is 26 atm. Answer is "a".

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