TITLE.PM5

444 ENGINEERING THERMODYNAMICS

dharm

\M-therm\Th9-3.pm5

= 4 × 0.745 loge 323

298

+ 6 × 0.653 loge 323

298

= 0.5557 kJ/K. (Ans.)

Note. ∆S may also be found out as follows :

∆S = ()mmNCO 2 + c

T

ve() T

mix log^2

1

= (4 + 6) × 0.6898 loge

323

298 = 0.5557 kJ/K.

Highlights

1. According to Dalton’s law :
   (i) The pressure of a mixture of gases is equal to the sum of the partial pressures of the constituents.
   (ii) The partial pressure of each constituent is that pressure which the gas would exert if it occupied
   alone that volume occupied by the mixture at the same temperature.


2. According to Gibbs-Dalton law :
   (i) The internal energy, enthalpy and entropy of a gaseous mixture are respectively equal to the sum of
   the internal energies, enthalpies and entropies of the constituents.
   (ii) Each constituent has that internal energy, enthalpy and entropy, which it would have if it occupied
   alone that volume occupied by the mixture at the temperature of the mixture.


3. The characteristic equation for mixture is given as :
   pV = nR 0 T
   where n = Number of moles of mixture, and
   R 0 = Universal gas constant.


4. Molecular weight (M) may be found out by using the following relations :
   M =
   n
   n

i

∑^ Mi and M =

1

m

M

fi

∑ i

where mf = mmi = mass fraction of a constituent.

5. The following condition must be satisfied in an adiabatic mixing process of perfect gas in steady flow :

T =

Σ

Σ

mc T

mc

i pci

i pi

=Σ

Σ

nC T

nC

i pii

i pi.

Objective Type Questions

Choose the Correct Answer :

1. In an ideal gas the partial pressure of a component is
   (a) inversely proportional to the square of the mole fraction
   (b) directly proportional to the mole fraction
   (c) inversely proportional to the mole fraction
   (d) equal to the mole fraction.


2. The value of the universal gas constant is
   (a) 8.314 J/kg K (b) 83.14 kJ/kg K
   (c) 848 kJ/kg K (d) 8.314 kJ/kg K.