Question

Prior to the discovery that Freon-12 (CF2Cl2) was harmful to the Earth’s ozone layer, it was frequently used as the dispersing agent in spray cans for hair spray, etc. Its enthalpy of vaporization at its normal boiling point of –29.2 oC is 20.25 kJ mol–1. Estimate the pressure that a can of hair spray using Freon-12 had to withstand at 40 oC, the temperature of a can that has been standing in sunlight. Assume that ΔvapH is a constant over the temperature range involved and equal to its value at –29.2 oC.

Answer 1

Answer:

The pressure that a can of hair spray using Freon-12 had to withstand at 40°C is 9,1 atm

Explanation:

It is possible to answer this question using Clausyus-Clapeyron formula:

$ln\frac{P_2}{P_1}=\frac{-{delta}H_{vap}}{R}(\frac{1}{T_2}-\frac{1}{T_1} )$

Where:

P₁ = 1atm -The vapour pressure of a liquid at its normal point is equal to normal atmospheric pressure, 1atm-

P₂ is the pressure we need to find.

ΔHvap = 20,25 kJ/mol

T₁ is the boiling point temperature, 243,95K (273,15 -29,2°C)

And T₂ sunlight temperature, 313,15K (273,15 +40°C)

Replacing:

$ln\frac{P_2}{1atm}=2,2062$

$\frac{P_2}{1atm}=e^{2,2062}$

$P_2} = 9,1 atm$

Thus, the pressure that a can of hair spray using Freon-12 had to withstand at 40°C is 9,1 atm

I hope it helps!