Chemistry, asked by Anonymous, 11 months ago

Can you please explain the difference b/w ispthermal and adiabetic process.....

Broad explanation required

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Answers

Answered by 8397987195
1

There are lots of places to look to find the difference between an isothermal process and an adiabatic process, including online, any introductory physics text, etc. But I think there are real reasons why these two processes are often confused with each other - and it is based on some fundamental misunderstandings.

In short: Isothermal simply means constant temperature. So any thermodynamic process that occurs at constant temperature is isothermal. An adiabatic process, on the other hand, is any process in which there was no heat exchanged with the object or system being described. The common misconception is that if no heat is exchanged with a system, doesn’t that mean its temperature was constant. Or conversely, if the temperature is constant, doesn’t that mean no heat was exchanged.

Let’s look at a couple of examples. (a) Suppose you place a 0°C ice cube on the table and just watch as it melts to become 0°C water. For the ice cube, that was an isothermal process since it’s temperature did not change as it went from solid to liquid form. But to melt, it had to absorb heat from the surroundings (the table and the air) - so the process was not adiabatic. (b) Suppose you place an ice cube into a cup of warm water, but the cup is a perfect thermal insulator. The ice cube may melt and the cold water may warm some as the warm water cools. So this process is not isothermal for either the ice or the water. But the system of ice and water did not absorb or give off any heat because the container was a perfect insulator. In that sense, the process was adiabatic.

But the question comes up most often with respect to thermodynamic processes on ideal gasses. That is, consider a piston-cylinder with a contained ideal gas. The state of that gas can be represented on a pV diagram - meaning a graph of the pressure of the gas as a function of its volume. Now let’s consider two cases in which the gas is expanded from a smaller volume to a larger one by having the piston move ‘outward’. Such a process can occur under several conditions. That is, the pressure can be held constant by adding heat to the system as the volume increases, and that will also increase the temperature. But the two cases I want to describe, because this is where a lot of confusion often occurs, are an isothermal expansion and an adiabatic expansion.

Isothermal expansion. If the volume expands while heat is being continually supplied to the contained gas in such a way to hold the temperature constant, it is called isothermal. But by the ideal gas law, pV=nRT, if the temperature is constant, as the volume increases the pressure has to simultaneously decrease. But if the temperature does not change, this process also guarantees that the internal energy of the gas is constant, since temperature is a measure of the internal energy of an ideal gas. But because the volume increased, there was work done by the gas in pushing the piston outward, so the first law of thermodynamics required that some heat be supplied to account for the amount of work done. So heat was absorbed, work was done by the gas, and there was no internal energy change.

Adiabatic expansion. If the same volume change occurred in this problem, but no heat had been added, it would be by definition adiabatic. But since the gas would have done work against the piston in this expansion, that energy would have had to come from somewhere. If there was no heat added to the system (as in the isothermal expansion), the only source of energy is the internal energy of the gas itself. So in the adiabatic expansion, there would be a decrease in internal energy - which meant a drop in temperature and hence a greater drop in pressure that occurred during the above isothermal expansion. So no heat was absorbed, work was done by the gas, and the internal energy of the gas decreased meaning the temperature dropped

Answered by HelpingHandJai
0

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