PHYSICAL CHEMISTRY IN BRIEF

CHAP. 4: APPLICATION OF THERMODYNAMICS [CONTENTS] 122

Solution

From equation (4.35) and from the assumption that the process is adiabatic it follows that

Wsh= ∆H= 49800−60705 =− 10905 J.

The obtained electric energy is−Wsh= 10905 J.

We calculate the volume work from (4.12)

Wvol=Wsh+p 1 V 1 −p 2 V 2 =−10905 + 7. 056 × 500 − 0. 6039 ×10000 =− 13416 J.

The volume workWvol=−13416 J cannot be used to the full to generate electric energy. Part

of this work, 13416−10905 = 2511 J, is consumed on pushing water steam through the turbine.

4.3.4 The Joule-Thomson effect

When a fluid passes through a barrier that offers resistance to it (e.g. porous glass or a throttle),
a change in temperature and pressure occurs in the fluid. This fact is called the Joule-Thomson
effect. If the fluid passing through the barrier does not exchange heat with the surroundings,
the transfer through the barrier is an isenthalpic process

H(T 1 , p 1 ) =H(T 2 , p 2 ), (4.36)

where the subscripts 1 and 2 indicate the state of the system before and after the barrier.

The ratio of the temperature and pressure difference,

T 2 −T 1

p 2 −p 1

, is called theintegral Joule-

Thomson coefficient.

Note: There is a connection between an engine with a steady flow of substance and the

Joule-Thomson phenomenon. If an engine with a steady flow operates adiabatically and

does not exchange any shaft work with its surroundings, equation (4.35) rearranges to

(4.36), i.e. the Joule-Thomson effect occurs.