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When an ideal gas with constant specific heat is throttled adiabatically, with negligible changes in kinetic and potential energies:

  • (a): ∆h = 0, ∆T = 0
  • (b): ∆h =0, ∆S > 0

The answer is supposed to be (b).

Please explain why answer (a) is wrong and (b) is correct.

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  • $\begingroup$ It is a bit confusing that there are 'negligible changes in kinetic energy' - usually in a throttling process internal energy is converted to kinetic energy, resulting in a decrease in temperature and an increase in velocity. $\endgroup$ – Carlton Aug 11 '15 at 14:38
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You answer this by applying first and second laws of thermodynamics.

First Law $$Q - W = \triangle E_{sys}$$ since throttling process is adiabatic and there is no mechanical work applied on or extracted from the system then $Q$ and $W$ = 0.

that leaves us with $\triangle E_{sys} = 0$, and for steady state system we can express $E$ as the sum of enthalpy, kinetic energy and potential energy:

$$E = m(h + v^2/2 + gz)$$ and since we have negligible changes in kinetic and potential energies we finally get:

$$\triangle h = 0$$

And finally, as we know from second law of thermodynamics that for an adiabatic process $\triangle S \geq 0$

(You should have a look on the throttling process on an h-s diagram of a reversed carnot cycle (refrigeration cycle).)

The temperature behavior in throttling process is governed by the Joule-Thomson coefficient $\mu$ $$\mu = \left ( \frac{\partial T}{\partial P} \right )_h $$

The Joule-Thomson coefficient is a measure of the change in temperature at constant-enthalpy process, if it equals zero then temperature remains constant, if greater than zero temperature decreases and if less than zero temperature increases.

However, for an ideal gas the enthalpy is a function of only the temperature $h = h(T)$, and since enthalpy remains constant in throttling processes and $\mu = 0$ (You can check Cengel's Thermodynamics book for the proof), the temperature also remains constant.

And so, all the choices are actually correct.

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  • $\begingroup$ Why isn't ∆T=0? I mean that ∆h=mc∆T. So, if ∆h=0 then ∆T should be zero, right? $\endgroup$ – M.Tarun Aug 11 '15 at 14:31
  • $\begingroup$ @M.Tarun Updated. $\endgroup$ – Algo Aug 12 '15 at 6:34

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