EXPERIMENTS TO MEASURE (u O – u)

I. EXPERIMENTS TO MEASURE (u O – u)

It is possible to measure the difference (u o – u) for real gases using the Washburn ex- periments (see Chapter 2). A mass m of high pressure gas stored in a tank A is discharged through s narrow tube C into the atmosphere at B ( Figure 12 ). The tank and the tube are main- tained in a constant temperature bath D. Recall the following relation from Chapter 2, during a short period of time “dt”.

(a) Since the discharged gas pressure is atmospheric,

m du + u dm = δQ – dm(h(T B ,P o ) + ke o ).

h(T B ,P o )=h o (T B ).

Ignoring the kinetic energy ke, Eq.(a) becomes d(mu) = δQ+ dm h o (T B ), i.e., m 2 u 2 –m 1 u 1 = Q + (m 2 –m 1 )h o (T B ). Where states (1) and (2) are initial and final states of tank A With u 2 =u 0 , m 2 =m 0 ,

(m 0 u 0 (T B )–m 1 u 1 (T 1 ,P 1 )) = Q + (m 0 –m 1 ) RT B and m 0 = v/v 0 , with m 1 = v/v 1

V/v o u o – V/v 1 u 1 = Q + (V/v o – V/v 1 )(u o + RT B ), i.e.,

u res

Figure 12. Washburn experiments (from A. Kestin, A Course in Thermodynamics, McGraw Hill, NY, 1979, p 262, Volume I. With permission.).

u o (T B )–u 1 (T B , P) = Q v 1 /V + (RT B v 1 /v o – RT B )=Qm 1 + (P o V/m 1 – RT B ). (b) With known Q, V, T B and P O we can determine the difference between the ideal and real gas

internal energies in this manner. The “P” in tank A can be altered and corresponding u 0 – u 1 can be determined from Eq. (b). Likewise, using the expression h = u + Pv,

h o (T B )–h 1 (T B , P) = Q m 1 – (P – P o ) V/m 1 .

The Washburn coefficient ( ∂u/∂P) T represents the slope of the difference (u o (T B ) – u 1 (T B , P)) with respect to pressure. The slopes for molecular oxygen and air, respectively, tend

to approach values of 6.51 and 6.08 kJ kmole –1 bar –1 as P →P o . Differentiating Eq. (60), we obtain the relation

∂P = –(3/2)(a/(bT 2 ∂v/∂P)/(v (1+b/v)). (c) Since v has a relatively large value (as P →P o ), neglecting higher order terms in v,

(T) – u(T,v))/ ∂(u 1/2

∂(u 2 o (T) – u(T,v))/ ∂P = –(3/2)(a/(bT ))( ∂v/∂P) T,v →∞ /v . Using the expression for ∂v/∂P given in Chapter 6 for RK equation, (

= 1/((a(2v+b)/T 1/2 2 2 ∂v/∂P) 2 T v (v+b) ) – RT/(v–b) ). As v → ∞, ∂v/∂P= –v 2 /RT.

(d) Therefore using Eq.(d) in Eq.(c), ∂(u o (T) – u(T, v))/

∂P = (3/2)(a/RT 3/2 ).

Rewriting this expression in reduced form and using the RK state equation relation for a =

2 0.4275 R 1.5 T

c /P c , ( ∂u C,R /

∂P 1.5 R ) = 0.6413/T R Using the values for a = 17.39 bar K 1/2 m 6 kmole –2 and R = 0.08314 bar m 3 kmole –1 K –1 for molecular oxygen, at 301 K, ∂(u o (T) – u(T,v))/ ∂P = 0.0601 m 3 kmole –1 or 6.01 kJ kmole -1 bar -1 while the experimental value is given as 0.0651 kmole -1

bar -1 .