Internal Energy

4. Internal Energy

If the volume of an ideal gas is expanded from v to v o while maintaining the same pressure as in a corresponding real gas, its internal energy u o is unchanged, since the intermo- lecular potential energy remains unaltered. Therefore, u o (T,v o )=u o (T,v) = u o (T). Furthermore, since u = a +Ts,

u(T,v) – u o (T) = a(T,v) – a o (T) + T {s(T,v) – s o (T,v)}. (60) Using Eqs. (47) and (53) in Eq. (60), we obtain the relation u(T,v) – u o (T) = u(T,P) – u

o (T) = RT ln {v/(v–b)} + {a/(bT )} ln {v/(v+b)} + RT ln {v/(v–b)} + (1/2){a/(bT 1/2 )} ln {v/(v+b)}.

Simplifying this expression. u(T,v) – u (T) = –(3/2)(a/(bT 1/2 o )) ln(1 + (b/v)).

(61a) The term u(T,v) –u o (T) or u(T,P) – u o (T) = u Res is the residual internal energy, and is

typically less than zero. Substituting for a and b in terms of critical properties, Eq. (60) as- sumes the form

0.08664 uC R = , −

u Res u (T) u(T,v) −

T 0.5 ln (1 +

) , (61b)

RTc

RTc

vR ′

where the difference u o (T) – u (T,v) represents the departure of the internal energy from that in

a corresponding ideal gas. Recall that

v R ´ = v/v c ´, v c ´ = RT c /P c .

These expressions form the basis for charts illustrating the behavior of –(u Res /RT c ) with respect to the reduced pressure, temperature, and specific volume.

a. Remarks We recall from Eq. (23)) that ∂(a/T)/∂(1/T) = u. Thereafter, dividing Eq. (47)

throughout by T and then differentiating with respect to (1/T), we obtain Eq. (61). The internal energy is generally higher in an ideal gas than in a corresponding real gas. Figure 8a presents a plot of u with respect to T at different specific volumes. At a specified temperature, as the volume is increased to a large value, u →u o (the dashed

line QR). The slope of line QR yields the value of c vo, while the slope of line AB

yields the value of c v . Further u o,R =u o +

vo dT.

The free volume per molecule is much larger in the gas or vapor phase than in the liq- uid phase due to the greater intermolecular distance. Therefore, since the vapor has a higher intermolecular potential energy (cf. the LJ diagram), the liquid has a lower in- ternal energy. The constant volume specific heat c v is obtained by differentiating Eq. (61a) with re- spect to temperature at a specified specific volume, i.e., The free volume per molecule is much larger in the gas or vapor phase than in the liq- uid phase due to the greater intermolecular distance. Therefore, since the vapor has a higher intermolecular potential energy (cf. the LJ diagram), the liquid has a lower in- ternal energy. The constant volume specific heat c v is obtained by differentiating Eq. (61a) with re- spect to temperature at a specified specific volume, i.e.,

saturated liquid to the saturated vapor state at constant temperature, then Eq. (62) suggests that

c v (T,v f )–c v (T,v g ) = 3/4 (a/(bT (3/2 ) ln((1 + b/v f )/(1+ b/v g )). (63) In general, since v f <v g ,c v (T,v f )>c v (T,v g ). Dividing Eq. (62) by R and using the rela-

tions for a and b in terms of the critical properties T c and P c , Eq. (62) assumes the form

v (T,v) – c v0 (T))/R = 3.7007/T R ln(1 + 0.08664/v R ´). (64) The variation of the reduced specific heat with P R and v R ’ with respect to volume is

(c

illustrated in Figure 9 . The RK based relation for c v vs. v for H 2 O at 593 K is plotted in Figure 10 . Instead of the RK equation, if one uses the following a hypothetical RK equation with

P = RT/(v–b) – a/(T n v(v+b)), then we can show that (c (T,v) – c (T))/R = (4.9342 n(n+1) /T n+1 v v0 R ) ln(1 + 0.08664/v R ´)). If -1 <n<0 (i.e., attractive forces increase with temperature), c v (T,v) < c v0 (T) and c v

could become negative! A negative specific heat implies that the temperature will in-

U (a)

(b)

Figure 8: (a) Illustration of the determination of the value of u at state B from a known value u o at state Q. (b Illustration of the determination of the value of h at state B from a known value h o at state Q.

0.6, Liquid l

Figure 9: (a) Values of reduced c vR with respect to P R for an RK fluid. The middle values are unstable and the upper values at specified values of P R and T R correspond to a liquid–like solution, while the lower values correspond to a gas–like solution. The value of c v for a liquid is always higher than that for

a gas. (b) Values of reduced c vR with respect to v R ´. The variation in the values is monotonic unlike

those in relation to P R .

Using Equation (38), ( ∂T/∂v) u = – (T( ∂P/∂T) v – P)/c v .

If the fluid state is defined by the RK equation, the numerator in the above relation is always negative, i.e., when an RK substance expands at constant internal energy, the temperature decreases. Such an adiabatic throttling process for a closed system is dis- cussed later. Recall that c p –c v =–T(

∂P/∂T) 2 /( ∂P/∂v). If (∂P/∂v) ≈ 0 (in Chapter 6 we discussed that for real gases the pressure decreases with a decrease in the specific volume over a

range of values of v when T < T c ), then c p –c v < 0, i.e., c p <c v . Therefore, the value range of values of v when T < T c ), then c p –c v < 0, i.e., c p <c v . Therefore, the value

a mixture can be obtained by using Eq. (61a). t. Example 20

Determine the internal energy of water at 250 bar and 600ºC if u o = 3302.7 kJ kg –1 at that temperature. The corresponding tabulated value of the internal energy from the steam tables ( Table A-4C ) is 3138 kJ kg –1 . Employ the RK equation in your solution.

Solution Since the critical properties for water are P c = 220.9 bar and T c = 647.3 K, P R = 1.132 and T R = 1.349, Therefore, upon applying the appropriate results from Chapter 6, v R ´ = 1.007, and

–u 0.5 C,R = 7.401/1.349 ln(1 + 0.08664/1.007) = 0.526, i.e., u – u = 0.526

. The difference with respect to the steam tables is 0.25 %.

× 0.461 × 647.3 = 157 kJ kg –1 , or u = 3146 kJ kg –1