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

the cold front replaces the warmer air at the ground level changing the wind to be breezy and denser.there is also an immediate drop in the temperature.

weight of any object is given by

w = m*g

where m is mass of the object and g is the acceleration due to gravity.

for earth the value of g is 9.8 m/s^2 so

w=117.6 n m=12 g=?

117.6=12g => g=9.8 m/s^2

which is the force of gravity on earth.