Phase changes: Solved problems

1. The latent heat of vaporization of water at 100oC is 539 kcal/kg. Calculate the boiling point of water at  600 mm Hg. The value of the universal gas constant is  1.98 cal/mole oK.

We use the integrated form of Clausius-Clapeyron equation:

P2l(T2-T1)
 2.3 log ----- = --------- 
P1RT1T2

Since R is given as cal/mole we need to transform the value of l by multiplying it with the molecular weight of water (18):

l = (539 kcal/kg) (18 kg/kmole) = 9702 kcal/kmole = 9702 cal/mole

Furthermore T = 100 + 273 = 373 oK. By inserting the known terms in the equation above we have that:

1.

6009702(T2 - 373)
 2.3 log ----- = ----- -------------
7601.98373 T2

2.

(T2 - 373)
 - 0.23 = 4900 -------------
373 T2

3.

(- 0.23) (373 T2)
------------------- - T2 = -373
4900

4.

-373  = -0.0175T2 - T2

5.

1.0175 T2 = -373

6.

T2 = 373/1.0175 =  366.5 oK

T2 = 366,5 - 273 = 93.5 oC

2. The vapor pressure of a substance is 723 mm Hg at 10oC and 769 mm Hg  at 11.2oC. Calculate the latent heat of vaporizzation of the substance.

We use the integrated form of Clausius-Clapeyron equation:

P2l(T2-T1)
 2.3 log----- = --------- 
P1RT1T2

By inserting the known terms in the equation above (after conversion of temperatures in kelvin) we have that

769l(284.2 - 283)
2.3 log ----- = ----------------------
7321.98 (283)(284.2)		 l 1.2
0.05 = ------------
159248

l = (0.05)(159248)/1.2 = 6635 cal/mole

3. 10 g of nitrogen (Mw = 28) are bubbled into liquid bromine (Mw = 160) at 720 mm Hg and 35 oC. Assuming that

Calculate the grams of bromine vaporized.

If nitrogen is saturated by bromine the partial pressure of bromine must match the vapor pressure of bromine at the given temperature (35oC). By knowing the vapor pressure of bromine at 25 oC (230 mm Hg) we can calculate, by using the integrated form of Clausius-Clapeyron equation, the vapor pressure at 35oC: 

P2l(T2-T1)P2(7300)(308 - 298)
 2.3 log----- = --------- = 2.3 log------ = -------------------- = 0.40
P1RT1T2230(1.98)(298)(308)
 
P20.40
 log------- = --------- = = 0.174
2302.3		 log (P2) = 0.174 + log (230) = 2.535
P2 = 343 mm Hg

According to Dalton's law

Pnitrogenmolesnitrogen
---------- = --------------
Pbrom.molesbrom.		 where

moles nitrogen = 10gr/mw = 10/28 = 0.357

Pnitrogen = 720 - 343 = 377 (total pressure less that exercised by bromine)

Pbromine = 343

Thus

3770.357
-------- = -------------- = 1.1
343molesbrom.
molesbrom. = (0.357)/(1.1) = 0.324 ==> (0.324) (mw) = (0.324) (160) = 52 grams

4. Calculate the fusion temperature for sulphur (atomic weight = 32) at 1000 atm. It is known that at 1 atm:

  • fusion temperature  = 387oC

  • change in volume during fusion = 0.041 l/kg

  • latent heat of fusion = 0.420 kcal/mole

    • Furthermore, assume that the difference between the volume of the solid and that of the liquid is constant, i.e., is the same at any considered pressure.

    In the case of sulphur, the difference between initial and final volume (Vf - Vi) can be considered constant. In other words at any pressure Vf - Vi is always the same. If also the latent heat is assumed to be constant, the C-C equation

    dPlfus
    ------- = ------------
    dtT(Vf - Vi)
    can be written as
    dP = k (dT / T) where k = l /(Vf - Vi)

    by integrating we get P2 - P1 = k 2.3 log (T2/T1). In order to calculate k, let's express the latent heat as (liter) (atm)/kg

    0.420 kcal/mole = 0.420/32 = 0.0131 Kcal/g ---> 13.1 Kcal/kg

    The coefficient to transform kcal in liter atm is 41.3, thus

    l = (13.1)(41.3) = 542 (liter) (atm)/kg

    Now we can calculate k: k = l /(Vf - Vi) = 542/(O.O41) = 13219 and then solve th C-C equation for T2:

    P2 - P1 = k 2.3 log (T2/T1)
    999 = (13221) (2.3) log (T2/387) = 30408 log (T2/387)
    log T2 - log 387 = 0.0328
    log T2 = 0.0328 + log 387 = 2.62
    T2 = 102.62 = 417 oK
    T2 = 417 - 273 = 144oC

    5. Calculate the volume and the enthalpy content of a steam at 120oC and 80% quality.

    From table of steam we have that

    specific volume of liquid = 0.0010603 m3/kg
    specific volume of steam  = 0.8919 m3/kg
    Enthalpy of liquid = 503.71 kJ/kg
    Entalpy of steam = 2706.3 kJ/kg

    Latent enthalpy is then 2706.3 - 503.71 = 2202.59 kJ/kg. The quality of steam is 80% this means that only 80% of latent heat has been supplied, i.e.,  0.8 x 2202.59 = 1762.072 kJ/kg. Therefore the enthalpy of steam is

    1762.072 (latent) + 503.71 (liquid) = 2265.78 kJ/kg

    Alternatively, using the formula previously described:

    Hx = (1 - Sq) Hl + SqHv = 1 - 0.8 (503.71) + 0.8(2706.3) = 2265.78

    Analogously we can calculate the volume

    Vx = (1- Sq)Vl + SqVv
    Vx = (1- 0.8) 0.0010603 + 0.8 (0.8919 ) = 0.7137 m3/kg

    Please note that if the volume of saturated liquid is neglected, only a small error occurs:

    Vx = SqVv = Vx = 0.8 (0.8919 ) = 0.7135 m3/kg

    6. How much heat will evolve when steam (100% quality) at 125 oC and 232.1 kPa is cooled to 120 oC at the same pressure?

    From steam tables we can deduce that:

    At 125 oC and 232.1 kPa water occurs as saturated vapor.

    At 232.1 kPa the boiling point temperature is 125oC thus at 120oC water is in liquid state.

    The enthalpy content of steam (125 oC and 232.1 kPa) is 2713 kJ/kg while the enthalpy content of water (120oC) is 503.69 thus the heat evolved after cooling is

    2713 - 503.69 = 2209 kJ/kg

    7. Water at 25 oC is added to a completely evacuated vessel (0 atm). Calculate the pressure in the vessel when the equilibrium liquid vapor is reached a 25 oC.

    The only source of pressure is the vapor in equilibrium with water. At 25 oC the vapor pressure of water is 3.17 kPa and thus the pressure in the vessel is 3.17 kPa.

    8. Water at 25 oC is added to vessel containing dry air at 1 atm. Assuming that no air escaped the vessel after the addition of water, calculate the pressure in the vessel when the equilibrium liquid vapor is reached a 25 oC.

    The total pressure is given by

    Pressure = Pair + Psteam = 101.3 kPa + 3.17 kPa (from steam table) = 104.47 kPa