0
$\begingroup$

Consider a system which is initially at a state $A$. Then it is moved to state $B$ absorbing $Q_h$ amount of heat from a heat reservoir at $T_h$. As the process is reversible so during heat exchange, the temperature of reservoir and the system will approximately be the same as finite temperature difference cause irreversibilities. Then if we come back to $A$ reversibly by the same path as of $A\to B$ but the direction being opposite, it dumps same heat $Q_h$ to a heat reservoir at temperature $T_h$.
So, $\oint_{A\to B}\frac{dq}{T}\;=\;\frac{1}{T_h}\oint_{A\to B}dq\;=\;\frac{Q_h}{T_h}$
Similarly, $\oint_{B\to A}\frac{dq}{T}\;=\;\frac{Q_h}{T_h}$
So $\frac{Q_h}{T_h}-\frac{Q_h}{T_h}=0$.
Hence, $\int_{A\to B\to A}\frac{dQ}{T}=0$.

Now suppose the system is at a state $A'$ such that the temperature of the system $T'$ (say, gas piston system) is significantly less than that of the temperature of reservoir $T_h$. If the sytem goes from $A'$ to $B'$ taking $Q_h$ amount of heat from the reservoir then the heat exchange occurs due to finite temperature difference between the system and the reservoir (source of irreversibility).

I have a doubt that in the case of irreversible process, what form does the $\int_{A'\to B'}\frac{dq}{T}$ takes?
As in that case temperature of the system changes abruptly from $T'$ to $T_h$ (final temperature due to heat transfer) and also the temperature in-between $T'$ and $T_h$ can not be even measured due to irreversible process, so I think $T$ can't be taken out from the integral.
I have a confusion that what form it will take. Please clarify the doubt.

Improve this question
$\endgroup$
10
  • $\begingroup$ In the irreversible case, the temperature doesn't jump abruptly from T' to Th. Only the boundary temperature in contact with the reservoir jumps to Th. And, in the integral, you are supposed to use Th instead of T. So, $$\Delta S>\frac{Q}{T_h}$$ $\endgroup$
    – Chet Miller
    Mar 26, 2021 at 18:32
  • $\begingroup$ @Chet, if we have a system of ideal gas at state A having temperature $T'$ and $Th$ at B. If we have a clamped piston (isochoric process), the heat we supplied increases its internal energy thus the temperature. If the difference between $T'$ and $Th$ is significant then the system gains heat irreversibly in fast manner. So boundary and system temperature changes abruptly, according to me. Can you please tell reasoning behind using $Th$ in the integral. I am not able to understand it. $\endgroup$
    – Iti
    Mar 26, 2021 at 18:42
  • $\begingroup$ Only the boundary temperature changes abruptly from T' to Th. The boundary temperature is the temperature of both the gas and the reservoir. The temperature change penetrates by conduction first only in the region close to the boundary, and then, as time progresses, further and into the interior. So, at short times, the very center of the gas does not even know that the boundary temperature has changed, and it is still very close to T'. Only at very long times is the entire gas at Th. See Transport Phenomena, Bird et al. $\endgroup$
    – Chet Miller
    Mar 27, 2021 at 9:52
  • $\begingroup$ See this thread that immediately precedes yours physics.stackexchange.com/questions/623955/… $\endgroup$
    – Chet Miller
    Mar 27, 2021 at 9:54
  • 1
    $\begingroup$ Yes, that is correct, even though the non-spatially-uniform temperatures in the interior of the system are less than Th throughout the process. And the integral gives the amount of entropy transferred from the surroundings to the system, rather than the total change in entropy of the system (which is greater). $\endgroup$
    – Chet Miller
    Mar 27, 2021 at 18:10

2 Answers 2

Reset to default
1
$\begingroup$

The entire gas does not heat up to Th all at once. First the boundary is at Th, and then the heat penetrates in further and further by (transient) heat conduction into the interior. So initially, the interior portion of the gas is still cold. It is like baking a turkey in the oven. First the outside heats up, and then the heat flows toward the middle.

The surface of the cylinder where the temperature is Th is the location where all the heat Q flows into the gas. So that is the temperature that we are supposed to be using in the Clausius inequality. In the end, the entire gas temperature is Th, so its change in entropy is the same as in the reversible process. But, in the reversible process, the boundary temperature is lower than Th throughout the process. So $$\Delta S>\frac{Q}{T_h}$$

Improve this answer
$\endgroup$
2
  • $\begingroup$ I have two doubts- i) I have studied that in the reversible process in which the system gains heat from the reservoir at Th, throughout the process, system as well as boundary temperature remains at Th(so there is no finite difference in temperature of boundary (and system) and the reservoir). Like in the first process of carnot engine cycle (isothermal expansion), system (with boundary) is at the same temperature as that of the reservoir But in last line, you have wrote "in the reversible process, the boundary temperature is lower than Th throughout the process". Can you please explain that. $\endgroup$
    – Iti
    Mar 27, 2021 at 3:19
  • $\begingroup$ ii) In irreversible process from A' to B', the system with its boundary is at temperature T'. The heat reservoir is at Th. So the boundary takes the heat from the reservoir and increase the boundary temperature to Th. Then from the boundary (which is at Th), heat gets transferred to the system. Is that you are saying? Because then we can use Th in the $int_{A\to B} dq/T$. But I have a doubt that when heat gets transferred from reservoir to boundary somewhat increasing its temperature (but less than Th), isn't in this intermediate step some heat also gets transferred to system simultaneously? $\endgroup$
    – Iti
    Mar 27, 2021 at 3:28
0
$\begingroup$

In Clausius theorem for second law of thermodynamics, the temperature in the integrand is the temperature of the reservoir, not the system:

$$ \oint \frac{\delta Q}{T_{surr}}. $$

Therefore, there exists no problem about measuring the system temperature in an irreversible process.

The $T_{surr}$ is not taken outside the integral, because the process might involve many reservois, as for the semi-reversible process. Many reservoirs with nearly contining change in temperaure are employed to slowly moving the system along the thermal process. $$ \oint \frac{\delta Q}{T_{surr}} = \sum_i \frac{\delta Q_i}{T_{surr, i}}. $$

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.