Question

Below the triple point (-56.6°C) the vapor pressure of solid CO2 (also known as “dry ice”) is given as: ln p = -3116/T + 16.01 The molar heat of melting of CO2 is 8330 J. Use just the data given here about CO2. State any assumptions you need to make to answer the following questions. (a) Calculate the vapor pressure exerted by the liquid CO2 at 25°C. (b) Explain why solid CO2 sitting on a laboratory bench evaporates rather than melts.

Answer 1

ln p = -3116/(298 Klevin) + 16.01 [where 25 degrees C = 273 + 25 = 298 Kelvin]

= -10.45 + 16.01

ln p = 5.55

p = 1.71 atm

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so from above part a, it is evident that at room temperature at 25 degrees, pressure of CO2 is slightly above than atmospheric pressure of 1 atm, so that means CO2 will be close to boiling at that temperature and 1.71 atm (more than usual atmospheric pressure] and hence would be in vapor state rather than in liquid state.