Microsoft Word - Cengel and Boles TOC _2-03-05_.doc

where Pi,2
yi,2Pm,2and Pi,1
yi,1Pm,1. Notice that the partial pressure Piof

each component is used in the evaluation of the entropy change, not the

mixture pressure Pm(Fig. 13–13).

Chapter 13 | 691

Partial pressurePartial pressure

of component of component i

at state 2 at state 2

Partial pressurePartial pressure

of component of component i

at state 1 at state 1

∆si°^ = = si°, 2 – si°, 1 – Ri lnln

Pi,i, 2

—––—––

Pi,i, 1

FIGURE 13–13

Partial pressures (not the mixture

pressure) are used in the evaluation of

entropy changes of ideal-gas mixtures.

EXAMPLE 13–3 Mixing Two Ideal Gases in a Tank

An insulated rigid tank is divided into two compartments by a partition, as

shown in Fig. 13–14. One compartment contains 7 kg of oxygen gas at 40°C

and 100 kPa, and the other compartment contains 4 kg of nitrogen gas at

20°C and 150 kPa. Now the partition is removed, and the two gases are

allowed to mix. Determine (a) the mixture temperature and (b) the mixture

pressure after equilibrium has been established.

Solution A rigid tank contains two gases separated by a partition. The pres-

sure and temperature of the mixture are to be determined after the partition

is removed.

Assumptions 1 We assume both gases to be ideal gases, and their mixture

to be an ideal-gas mixture. This assumption is reasonable since both the

oxygen and nitrogen are well above their critical temperatures and well below

their critical pressures. 2 The tank is insulated and thus there is no heat

transfer. 3 There are no other forms of work involved.

Properties The constant-volume specific heats of N 2 and O 2 at room temper-

ature are cv,N 2 
0.743 kJ/kg · K and cv,O 2 
0.658 kJ/kg · K (Table A–2a).

Analysis We take the entire contents of the tank (both compartments) as

the system. This is a closed systemsince no mass crosses the boundary dur-

ing the process. We note that the volume of a rigid tank is constant and thus

there is no boundary work done.

(a) Noting that there is no energy transfer to or from the tank, the energy

balance for the system can be expressed as

By using cvvalues at room temperature, the final temperature of the mixture

is determined to be

(b) The final pressure of the mixture is determined from the ideal-gas relation

where

Nm
NO 2 NN 2 
0.2190.143
0.362 kmol

NN 2 

mN 2

MN 2

4 kg

28 kg>kmol

0.143 kmol

NO 2 

mO 2

MO 2

7 kg

32 kg>kmol

0.219 kmol

PmVm
Nm Ru Tm

Tm
32.2°C

1 4 kg 21 0.743 kJ>kg#K 21 Tm20°C 2  1 7 kg 21 0.658 kJ>kg#K 21 Tm40°C 2
0

3 mcv 1 TmT 124 N 2  3 mcv 1 TmT 124 O 2 
 0

0 
¢U
¢UN 2 ¢UO 2

EinEout
¢Esystem

O 2

7 kg

40 °C

100 kPa

N 2

4 kg

20 °C

150 kPa

Partition

FIGURE 13–14

Schematic for Example 13–3.

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