TITLE.PM5

428 ENGINEERING THERMODYNAMICS

dharm

\M-therm\Th9-2.pm5

M =

1

2

2

2

2

2

2

() () ()O ()

O

N

N

CO

CO

CO

CO

m

M

m

M

m

M

m

M

f ++ +f f f

=

1

01

32

07

28

015

44

005

28

... .++ +

=^1

0 03332.

= 30 kg/kg mole

Considering 1 kg of the mixture

n =

m

M

=^1

30

= 0.033

Now mole fraction xi =

n

n

i

Σ i

=

n

n

i = mM

n

ii/

∴ xO 2 =

(./ )

.

01 32

0033

= 0.094 ; xN 2 =

(. / )

.

07 28

0 033

= 0.757

xCO 2 = (. / )

.

015 44

0 033

= 0.103 ; xCO =

(. / )

.

005 28

0 033

= 0.054

(i)Partial pressures of the constituents :

PO 2 = xO 2 × p = 0.094 × 1.3 = 0.132 bar. (Ans.)

PN 2 = xN 2 × p = 0.757 × 1.3 = 0.984 bar. (Ans.)

PCO 2 = xCO 2 × p = 0.103 × 1.3 = 0.1334 bar. (Ans.)

PCO = xCO × p = 0.054 × 1.3 = 0.0702 bar. (Ans.)

(ii)Gas constant of mixture, Rmix :

Rmix =

R

M

(^0) =^8314
30
.
= 0.277 kJ/kg K. (Ans.)
+Example 9.10. A mixture of ideal gases consists of 4 kg of nitrogen and 6 kg of carbon
dioxide at a pressure of 4 bar and a temperature of 20°C. Find :
(i)The mole fraction of each constituent,
(ii)The equivalent molecular weight of the mixture,
(iii)The equivalent gas constant of the mixture,
(iv)The partial pressures and partial volumes,
(v)The volume and density of the mixture, and
(vi)The cp and cv of the mixture.
If the mixture is heated at constant volume to 50°C, find the changes in internal energy,
enthalpy and entropy of the mixture. Find the changes in internal energy, enthalpy and entropy
of the mixture if the heating is done at constant pressure.
Take γ : for CO 2 = 1.286 and for N 2 = 1.4.
Solution. (i) The mole fraction of each constituent :
Since mole fraction, xi =
n
n
i
Σ i
∴ xN 2 =
4
28
4
28
6
44

• =
  01428
  01428 01364
  .
  ..+ =
  01428
  02792
  .

. = 0.511. (Ans.)