TITLE.PM5

422 ENGINEERING THERMODYNAMICS

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\M-therm\Th9-1.pm5

(i)Gas constant for air :

Now using the equation,

R =

m

m

i

∑ Ri, we have ...(ii)

R = 0.2314 × 0.2598 + 0.7553 × 0.2969 + 0.0128 × 0.2078 + 0.0005 × 0.1889

= 0.0601 + 0.2242 + 0.00266 + 0.00009 = 0.2871 kJ/kg K

Hence gas constant for air = 0.2871 kJ/kg K. (Ans.)

(ii)Apparent molecular weight :

Now from eqn. (i), we have M =

8 314

0 2871

.

.

= 28.96

Hence apparent molecular weight = 28.96. (Ans.)
Example 9.3. Following is the gravimetric analysis of air :
Constituent Percentage
Oxygen 23.14
Nitrogen 75.53
Argon 1.28
Carbon dioxide 0.05
Calculate the analysis by volume and the partial pressure of each constituent when the total
pressure is 1 bar.
Solution. We know that the analysis by volume Vi/V, is the same as the mole fraction
ni/n. Also ni = mi/Mi ; therefore considering 1 kg of mixture and using a tabular method, we have

Constituent mi Mi ni = m

M

i

i

n

n

i × 100% = V

V

i × 100%. (Ans.)

Oxygen 0.2314 32 0.00723 0 00723 100.0 03453. × = 20.94%. (Ans.)

Nitrogen 0.7553 28 0.02697 0 02697 100

0 03453

.

.

× = 78.10%. (Ans.)

Argon 0.0128 40 0.00032 0 00032 100

0 03453

.

.

× = 0.93%. (Ans.)

Carbon dioxide 0.0005 44 0.00001 0 00001 100

0 03453

.

.

× = 0.03%. (Ans.)

n = Σ ni = 0.03453

Also,

p

p

i = V

V

n

n

ii=

∴ pi = n

n

ip hence

For O 2 , PO 2 = 0.2094 × 1 = 0.2094 bar. (Ans.)

For N 2 , PN 2 = 0.7810 × 1 = 0.7810 bar. (Ans.)

For Ar, PAr = 0.0093 × 1 = 0.0093 bar. (Ans.)

For CO 2 , PCO 2 = 0.0003 × 1 = 0.0003 bar. (Ans.)