Carbon tetrachloride (CCI_4) is a carcinogen, and so chemists are interested in
ID: 488225 • Letter: C
Question
Carbon tetrachloride (CCI_4) is a carcinogen, and so chemists are interested in its properties under laboratory conditions. The normal boiling point and enthalpy of vaporization for carbon tetrachloride are T_b* = 76.8 degree C and delta H degree_ vap = 30.01 kJ/mol. Based on this information, estimate the vapor pressure of CCI_4 at T = 20.0 degree C. Give your estimated vapor pressure in units of torr. The density of solid water at T = 0.00 degree C is rho = 0.9167 g/cm^3, while that of liquid water at the same temperature is rho = 0.9998 g/cm^3. The enthalpy of fusion of water is delta H degree_ fus = 6008. J/mol. Based on this information, estimate the value for the melting point of water at p = 1000. atm. "Synthetic air" is a mixture of nitrogen (N_2) and oxygen (O_2) with X(N_2) = 0.80. Find delta G degree_ mix, delta H degree_ mix, and delta S degree_ mix when 5.00 moles of synthetic air is prepared from pure nitrogen and pure oxygen gas at T = 20.0 degree C. You may assume ideal behavior in this problem.Explanation / Answer
Clasius claypron equation is
ln(P/P0) = Hvap/R (1/T0 - 1/T)
where, Hvap = Heat of vaporization
P0 = vapours pressure at known temperatureT0
P = Vapour pressure at unkown Temperature T
R = gas constant , 8.314 J/Kmol
Hvap for CCl4 = 30.01KJ / mol = 30010J/mol
boiling point of CCl4 = 78.6°C = 351.75K , Which is T0
T= 20°C= 293.15K
At boiling point vapour pressure reach atmospheric pressure
So , P0 = 1 atm
P= ?
Applying the values in clasius claypron equation
ln(P/1atm)=30010(J/mole)/8.314(J/Kmol)×((1/351.75K)-(1/293.15))
2.303logP - 2.303log1= 3609.57 × ( 0.002843 - 0.003411)
2.303logP = -2.050
logP = -2.050/2.303 =-0.890
P2 =1×10^-0.890
=0.13atm
TTherefore, Vapour pressure of CCl4 at 20°C is 0.13atm
Therefore, at 20°C vapour pressure of CC4 is 0.13atm