Thermodynamics and an Introduction to Thermostatistics 3, Exercises of Chemistry

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  2 2 2 1 2 1 Problem 1 Problem 2






   





  
  )        
   Problem Set 3 Solutions 3.20 MIT Professor Gerbrand Ceder Fall 2001 top bottom top bottom bottom top air top top
gh P top top bottom top top bottom top bottom ln =
1 ln =
1 1 = + ln + = ln 1 + 1 ln (1 + ) =
1 1 =
1 1 = = = ( ) =
( ) 1 1 = 10 = 28 8 = 1000 4 184 8 314 298 10 0 0288 1 368 1 378 = 2 6 ln (1 ) =
ln =
1 1 d P H R d T P P H R T T P P
gh P P
gh x
gh P H R T T h H P
g T T P
PV M nRT M RT MW h H RT g MW T T g m/s ,MW . g/mol, C h cal/mol . J/cal . J/molK K m/s . kg/mol K K h . km d P d /T H R P P H R T T There are two steps to solving this problem (1) What is the pressure difference in terms of the height? (2) What is the height of the hill? Use the following form of the Clapeyron equation (this is valid for equilibrium between an ideal gas and a solid or liquid) Which in this case can be written in terms of the pressure at the top and bottom of the hill, and the respective boiling points Now we write the pressure at the bottom in terms of the pressure at the top Substituting this, If we expand we get, Assuming that the air is an ideal gas, we can relate the density and pressure to the temperature using the ideal gas law where (MW) is the molecular weight of the air. We now have Plugging in some numbers ( and we are dealing with air at 25 ) (a) Integrate the Clausius-Clapeyron Equation 1 Fall 2003        Problem 3    
  
   1 1 2 2 2 4 4 3 3 2 1 2 1







   
→

solid liquid total mix vap total vap b vap vap vap P T K, T K P , . P . P P . , T , . T . . T T K P T . . . P . . P nRT V . . . P P . n PV RT . . . . . . . H H H Q H H T . T K H . P P H R T T H = 1 = 1620 = 1000 ln = 156 000 8 314 1 1000 1 1620 = 7 6 10 = 13 049 19 314 = 156 000 8 314 1 1 1620 1 47 = 550 5 = 375 ln( ) = 30671 + 12 5 = 30671 1500 + 12 5 = 7 95 = 3 54 10 = 35 8 = = 1 107 87 (8 314) (1500) = 115 61 = = 35 8 = = 35 8 8 314 1500 = 2 87 10 2 87 10 (107 87 ) = 0 31 0 31
=
+
=
=
ln(1) = 30671 + 12 5 = 2454
ln =
1 1
For this example, atm (at boiling), and atm (b) At the melting point of Li, First we can find what the vapor pressure over the liquid would be under these conditions atm Pa If all of the silver was in the vapor phase (1 gram) Pa Hence, there must be both liquid and vapor present Pa Now we can find the moles in the vapor phase moles Grams of vapor moles g/mol g So the fraction in the vapor phase is (b) In general the amount of heat necessary would be the sum of the mixing enthalpy and the enthalpy of vaporization However, since we are told to assume the solution is ideal, the enthalpy of mixing is zero. (This is related to the assumption that the mixing species do not interact) Thus Since we were not given the enthalpy of vaporization or the boiling point in the data, we must find these quantities. To find the boiling point, use the given equation for the vapor pressure and find the temperature where the pressure is equal to 1 atm. To find the heat of vaporization, we can use the integrated form of the Clapeyron equation for an ideal gas. We know the temperature and pressure at two points (from the given equation for vapor pressure) and we can then calculate Choosing T= 1500 and 1600 K, the value calculated for is 255.kJ/mole. (If you look this up you get 251 kJ/mole) Now we can get the enthalpy of vaporization at 1500K, which can be done using a loop, since we know enthalpy is a state function 2