thermodynamics

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

3-29. The specific heat capacitycv of solids at low temperature is given by the equation

Cv= A()³

A relation known as the Debye T₂ lae. The constant equal to 19.4 x 10⁵ J kilomole-1 K^-1 and ᶿ is the Debye. Temperature, equal to 320  K for NaCl. What is the molal specific heat capacity at constant volume of  NaCl (a) at 10 K (b) at 50 K? (c) how much heat is required to raise the temperature of 2 kilomoles of NaCl from 10K to 50K at constant volume ? (d) what is the mean specific heat capacity at constant volume over this temperature range ?

ANSWER:

(a). The calculation is straight forward, inserting the temperature to the given, equation.

Cv (10 K) = 19.4 x 10⁵ J kilomole^-1 K^-1 x  ()

= 59. 204 J kilomole^-1 K^-1

Cv (50 K) = 19.4 x105 J kilomole-1 K^-1 x   ()

= 7400.5 J kilomole^-1 K^-1

(b) Using the definition of

                                      Cv= A()_v

                                       dq = C_v dT  integrate the above equation

                                      q = ³ Dt

                                      q = T⁴

                                       q= 9.236 x 10⁴

Q = 2q = 1,85 x10⁵

(c) the mean specific heat capacity can be calculated by dividing the sum of (Cv (10 K) + Cv (50 K) / 2by = 3729. 825 kilomole^-1 K^-1

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

3-33. A fictional metal of atomic weight 27 has a density of 3000 kg m^-3. The heat of fusion is 4 times 10^5 J kg^-1 at the melting point (900 K), and at the boiling point (1300 K) the heat of vaporization is 1.20 times 10^-7 J kg^-1. For the solid, c_P can be given by 750 + 0.5 T in J kg^-1 K^-1 and in the liquid c_p is 1200 J kg^-1 K^-1 independent of temperature.. (a) Calculate the heat of sublimation of the metal sample of the previous problem assuming that the heats of vaporization and fusion are independent of temperature and pressure. (b) Calculate the change of internal energy of the metal sample upon melting. (c) Calculate the change of internal energy of the metal sample upon vaporizing. Justify the approximations which must be made.

ANSWER:

(a)  For sublimation, the metal should be vaporized

So, heat required front 300K to 1200K  is 13,900 Joule. Now, heat required for sublimation :

= 13,900 + 0.01 x 1,200  x ( 1,300 -1,200) + 0.01 x 1,2 x10⁷

= 13,900 + 1,200 + 120,000

= 135,100 Joule

= 135,1 KJ is heat of sublimation for the sample

 

(b)  Change internal energy

ΔU = ddQ.

So, :   ΔUm = m x L

          ΔUm = 0,01 x 4x10⁵

ΔUm = 4,000 Joule

 

(c)   Change of internal energy upon vaporizing

(ΔU)_v = m x L_uap

(ΔU)_v = 0.01 x 1,2x 10⁷

(ΔU)_v = 120,000 Joule

Assuming no change in K_E or P_E no any external work is performed

The heat given, the all the heat directly goes to the internal energy of the modal.