Ideal Gas Mixture

6. Ideal Gas Mixture

a. Volume Since gases are ideal, there are no intermolecular forces. Hence an ideal gas ˆv k = v k , and, hence,

V(T, P, N 1 ,N 2 , ...) = Σ k v k (T, P) N k .

Using the ideal gas law for each component, V(T, P, N 1 ,N 2 ,...) = Σ k ( R T/P)N k =V 1 +V 2 + ... = N( R T/P).

(41) Equation (41) is a representation of the law of additive volumes. For a mixture the relation

assumes the form

(42) Equation (42) suggests that

PV = N R T = (N 1 +N 2 + ...) R T.

(43a) Therefore,

( ∂V/∂N k ) T, P, N N N 1 , 2 ,..., jk ≠ ,..., N K = ˆv k = R T/P = v k .

v ig = Σ k X k ˆv k = Σ k X k v k (43b)

b. Pressure Another form of Eq. (42) is

P=N 1 R T/V + N 2 R T/V + ... = p 1 (T,V,N 1 )+p 2 (T,V,N 2 )+…. (44) If each component alone occupies the whole volume, the pressure exerted by component k is

p k =N k R T/V (which is called component pressure and is the same as the partial pressure for ideal gases). Then the pressure exerted by the mixture

P= Σ k p k (T,V,N k ), (45) which is also known as Dalton’s law of additive pressure.

e. Example 6 Partially combusted products consist of propane (C 3 H 8 ): 0.5, O 2 : 2.5, CO 2 : 1.5, and

H 2 O: 2 kmoles at T = 298 K and P= 1bar. Determine the partial molal volume of CO 2 using the ideal gas mixture model at 1 bar and 298 K.

Solution The mixture consists of 0.5, 2.5, 1.5 and 2 kmole of C 3 H 8 ,O 2 , CO 2 and H 2 O, respec- tively.

V=( R T/P) Σ k N k , and

ˆv CO 2 =( ∂V/∂N CO 2 ) T, P, N N N 1 , 2 ,..., jk ≠ ,..., N K = R T/P = v

= 0.08314 ×298÷1 = 24.78 m 3 kmole –1 , and is the same for all species.

Thus V = v N= 161.07 m 3 .

If a room contains the mixture at T and P, we can hypothesize that 0.5 kmole (i.e.,

26 3 3 3 ×10 molecules) occupies 12.39 m while O

2 occupies 61.95 m .

c. Internal Energy The internal energy of a system is the combined energy contained in all of the mole- cules in the system. In ideal gases the internal energy of a species in a mixture equals its en- ergy in a pure state at the temperature and total pressure of the mixture. Therefore,

U 0 (T,N) = Σ k u k,0 (T) N k . (46) Eq. (46) is known as Gibbs-Dalton (GD) law Dividing throughout by N

u 0 (T,X) = Σ k X k u 0 (T)

d. Enthalpy Since

(48) as before,

H 0 =U 0 + PV 0 ,

H 0 (T,N) = Σ k h k,0 (T)N k

h k (T) = u k,0 +P v k = u k,0 + R T, ideal gas (49a) Similarly

h 0 (T) = Σ k X k h k,0 (T)

(49b)

e. Entropy Two gases A and B are contained in a chamber on two sides of a rigid partition in volumes V A and V B at specified values of temperature and pressure. The partition is removed and the gases are allowed to mix in a volume V, but the pressure and temperature are un- changed. For an ideal gas

dS = N B c v,o dT/T + N B R (dV/V).

A change in volume occupied by, say, species B from V B to V (e.g., if there are two adjacent rooms with O 2 and N 2 in each room and if the partition between them is removed, the N 2 now occupies a larger volume making more quantum states available for energy storage) results in an entropy change at same temperature. Then,

(T,V, N B, mix B )–S B (T, V B ,N B )=N B R ln(V/V B ).

The volume V = N R T/P and V B =N B R T/P, i.e., S–S B =N B R ln(V/V B ). The volume V = N R T/P and V B =N B R T/P, i.e.,

B mix (T,V, N B )–S B (T,V B ,N B ) =N B R ln(N/N B ) = –N B R ln(X B ). Dividing by N B ,

ˆs B (T, P, X B )– s B (T, P) = – R ln(X B ).

Further,