Question

Portable lanterns and stoves used for camping and backpacking often use a mixture of C5 and C6 hydrocarbons known as "white gas." White gas is easy to transport but stoves that burn white gas are harder to light in the cold. The figure below shows the structure of pentane, a typical C5 alkane, its normal boiling point is 36.0 oC and its heat of vaporization is 27.6 kJ/mol. Determine the vapor pressue of pentane on a cold autumn morning when the temperature is 1.50 °C.

Answer 1

Boiling point of a liquid is a temperature at which its vapor pressure is equal to atmospheric pressure. (1 atm or 760 mmHg).

Here, two different temperature are and heat of vaporization is given.

Using Clausius-Clapeyron Equation,

ln P₂ / ln P₁ = (delat H_vap / R) x ( 1/T1 - 1/ T₂)

were, P₂  and  P₁ are vapor pressure at temperature T₂ and T1 respectively.

H_vap = heat of vaporization.

  R = gas constant = 8.314 J/ mol . K

P₁  = 1 atm = 760 mmHg

P₂  = x (To find)

T₁ = 36.0  C = 309 K

T₂ = 1.50 C = 274.5 K

detla H = 27.6 KJ / mol

Putting all equation in the equation,

ln 1 - ln x = [ 27.6 / (8.314 x 10 ^-3) ] x [ (1/309) - (1/274.5)]

ln x = - 1.35026 atm

x = antiln (- 1.35026)

x = 0.2592 atm

P₂ = 0.259 atm .

Vapor pressure of Pentane at 1.50 C is 0.259 atm is the final answer.

As pressure and temperature are directly proportional, here temperature decreased and hence the vapor pressure too.