0
$\begingroup$

In isothermal process the work produced in a cyclic transformation is the sum of the small quantities of work $ \delta W$ produced at each stage of the cycle. This description of work I picture, for instance, as a cylinder with a piston filled with a gas, state of which is $p_1, V_1$. Then some weight is placed on the top of cylinder which compresses the gas to state $(p_2, V_2)$. The work destroyed by this process is equal to $W = P_0(V_1-V_2)$ and $P_0 > P_2 $, $P_0$ is pressure on the piston produce by the weight. Right?

So I have few question about this: Is the destroyed work reflected by change in temperature in the surrounding because the rapid compression produces increase of thermal energy inside the cylinder which is immediately sucked to surroundings (to satisfy the isothermal condition)? The physical effect of increase of thermal energy inside the cylinder (can I say for infinitesimal amount of time?) is caused by rapid movement of piston hitting (and thus accelerating) the molecules of the gas, hence the more massive the weight, the faster the change in state and the more "brutal" acceleration of gas molecules caused by collision with the piston?

The same description goes for heat: The heat withdrawn from the surround­ings in a cyclic transformation is the sum of the small quantities of heat $\delta Q$ withdrawn at each stage of the cycle. But I can't quite picture, maybe I'm misunderstanding the concept of heat. How can a heat be lost (analogous to work destroyed)? Can you give me an example of reversible and irreversible heat process?

$\endgroup$
0
$\begingroup$

Clarifications

Thermodynamics is tricky. We have to think carefully about what reversible means.

First of all, there are a couple of issues I see. First, there is no such thing as an isothermal cycle (unless you just retrace the same isotherm, but this is a pretty useless cycle). Second, work is not "destroyed", because work is not "stuff". Just like heat, work is the exchange of energy between two systems, or between a system and its environment. Third, if you place a weight on the piston, rapidly compressing the gas inside the cylinder doing work $W = p_0(V_1 - V_2)$ where $p_0 = m g/A$ ($A$ is the cross-sectional area of the piston) in the process, this process is not even quasi-static, which means that during the process, $p$ and $T$ (among other quantities) are not even well defined for the entire gas.

An isothermal process can't be fast

The fourth is that an isothermal process can't be fast, so your idea of placing a weight on the piston to rapidly compress the gas doesn't create an isothermal process. In order for a system to undergo an isothermal process, it must be held in contact with a reservoir of the same temperature, and whatever work is done must be done slow enough that there is enough time for heat to flow in or out in order to offset the energy gain/loss due to the work, i.e. $\delta Q = -\delta W$ (if we are defining $W$ so that it is the work done on the gas, i.e. $\delta W = -pdV$).

I usually picture the following situation (for simplicity, let's consider an isothermal compression). We have an ideal gas inside a cylinder-piston assembly in thermal equilibrium with the room in which it is placed; the massless piston is free to move. We place grains of sand one by one on the piston so that it moves downward incrementally. Each grain of sand does work $\delta W$ on the gas (via the massless piston), and since this amount of work is essentially infinitesimal, the infinitesimal increase in temperature caused by the small compression is immediately offset by a an infinitesimal amount of heat flow $\delta Q$ from the gas to the room.

Now, since the system gains the same amount of energy as it loses, this means that the system's environment gains the same amount of energy as it loses, and so the environment's temperature doesn't change either. What's important here is that the system and the environment exchange energy via both work and heat, we can't say that work is "destroyed". Rather, the environment loses energy via work, and the system gains that same amount of energy via work; the system loses the same amount of energy via heat, and the environment gains the same amount of energy via heat.

After many grains of sand have been placed on the piston (say, a mass $m$ of sand), the piston has been compressed from an initial volume $V_1$ to a final volume $V_2$, and since the temperature is the same, the pressure accordingly increases from $p_1$ to a final pressure of $p_2$. (We are assuming that the amount of gas inside the piston is constant.) The amount of work done on the gas in this process is

$$ W = nRT\ln\left(\frac{V_1}{V_2}\right) $$

which can be found using $\delta W = -pdV$ and the ideal gas law. This is exactly the "adding up" of little $\delta W$'s you were talking about. The reason this is not equal to $W = p_0(V_1 - V_2)$ where $p_0 = m g/A$ ($A$ is the cross-sectional area of the piston) is because not all of the sand (not all of the mass $m$) is on the piston the entire time.

Now, to your specific questions

Is the destroyed work reflected by change in temperature in the surrounding because the rapid compression produces increase of thermal energy inside the cylinder which is immediately sucked to surroundings (to satisfy the isothermal condition)?

We addressed this above: if the process is isothermal, then the environment loses as much energy by pushing on the system as it gains from heat flow from the system. It is important again to note it does not make sense to say that work is "destroyed", because work is an exchange of energy between two systems.

The physical effect of increase of thermal energy inside the cylinder (can I say for infinitesimal amount of time?) is caused by rapid movement of piston hitting (and thus accelerating) the molecules of the gas, hence the more massive the weight, the faster the change in state and the more "brutal" acceleration of gas molecules caused by collision with the piston?

This is correct! The piston moves downward, collides with ideal gas molecules, and the molecules gain a little bit of kinetic energy because of this collision, raising the temperature. If the mass is larger, than the change in the piston's height is more rapid, and the collisions are more "brutal". If the process is rapid, then the gas at the top of the cylinder gets compressed first, which means that there is no well-defined pressure for the entire gas. This is why the process is not quasi-static.

Reversible process with heat flow

There is only one process that is reversible in which heat flows: an isothermal one. This is because heat flow across any finite temperature difference is irreversible. The system and the environment must be at the same temperature in order for them to exchange heat reversibly. We have discussed this above in the context of the isothermal process.

An arbitrary process (say, a constant volume process) can be made to be reversible if we bring the system into contact with a sequence of reservoirs, each of which has a temperature infinitesimally different than the last, and the system is allowed to exchange heat approximately isothermally with each of the reservoirs. This is, of course, impossible to do in practice, and therefore:

All real processes are irreversible

Consider a constant volume process (an ideal gas inside a closed, rigid container): the container is placed in boiling water, so that there is a finite temperature difference between the gas and the water. Then, heat flows from the water to the system irreversibly. Again, heat is not "lost" in the sense that heat is the exchange of energy between the gas and the water. The water instead loses thermal energy because some of that energy has gone into the gas, and the exchange process is known as heat in this case.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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