Butane has a heat of vaporization of 22.44 kJ/mol and a normal boiling point of -0.4 degrees C. A 250mL sealed flask contains 0.5g of butane at 25 degrees C.

Question

Butane  has a heat of vaporization of 22.44 kJ/mol and a normal boiling point of -0.4 degrees C. A 250mL sealed flask contains 0.5g of butane at 25 degrees C.

How much butane is present as a liquid?

If the butane is warmed to 25 degrees C, how much is present as a liquid

DrBob222 · Answer

For the second part, use the Clausius-Clapeyron equation.
P butane at normal boiling point will be 760 torr. Solve for P @ 298, then substitute P into PV = nRT and solve for n. Compare with the number of moles (0.5 g/molar mass butane) present in liquid form.

For part a, I think you can use PV = nRT as is (P at -0.4 C is 1 atm), then compare n from this with n of 0.5 g butane.