Propane has a normal boiling point of −42.0∘c and a heat of vaporization (δhvap) of 19.04 kj/mol. what is the vapor pressure of propane at 25.0∘c?

Answer : The vapor pressure of propane at is, 7067.24 torr

Solution : Given,

Initial temperature =    

Final temperature =

   (1 kj = 1000 j)

At normal boiling point, the vapor pressure is equal to 760 torr.

So, initial pressure = 760 torr

Using Clausius Clapeyron equation,

where,

= vapor pressure at temperature

= vapor pressure at temperature

= enthalpy of vaporization

R = Gas constant = 8.31 j/Kmole

Now put all the given values in this formula, w get

Therefore, the vapor pressure of propane at is, 7067.24 torr

specific gravity is the density of a substance divided by the density of water. since (at standard temperature and pressure) water has a density of 1 gram/cm3, and since all of the units cancel, specific gravity is usually very close to the same value as density

so "c" is the answer

plz give brainliest

answer; protein;