Question

hi, can someone answer this question for me?

The triple point of carbon dioxide is p=518500 Pa, T=216.58 K, and its molar enthalpy of sublimation is 26.1 kJ mol-1. Calculate the sublimation temperature, expressed in K and correct to 2 decimal places, of carbon dioxide at a pressure of 441 kPa.

Answer 1

Sol . As Clausius - Clapeyron equation ,

ln (P2 /P1 ) = - (deltaH / R) ( (1/T2) - (1/T1) )

where P1 and P2 are initial and final pressures respectively

T1 and T2 are initial and final temperatures respectively

deltaH = Molar enthalpy of sublimation

R = Gas constant = 8.314 J / K mol

Now ,

P1 = 518500 Pa  

P2 = 441 kPa = 441000 Pa

T1 = 216.58 K

T2 = ?

deltaH = 26.1 KJ / mol = 26100 J/mol

So ,

ln(441000 / 518500 ) = - (26100/8.314) ( (1/T2) - (1/216.58) )

ln(0.85053) × ( - 8.314 /26100 ) = (1/T2) - (1/216.58)

0.0000515709 = (1/T2) - (1/216.58)

(1/T2) = 0.0000515709 + (1/216.58) = 0.0046688024

So , T2 = 214.19 K

Therefore , Sublimation temperature is  214.19 K