https://www.youtube.com/watch?v=nkOtOMNS5bQ&list=PLbKSbFnKYVY28nQHsrW-YPw2OZDx1OHug&index=6

Khan academy vid says go ahead and use delta U =0 on the MCAT

https://preview.redd.it/rm1relqgczt41.png?width=1764&format=png&auto=webp&s=675a2abad6dacfc9c12016a01d31c2f857289d40

And one guy on stack exchange defends it:

https://preview.redd.it/0r6hm0aiczt41.png?width=1482&format=png&auto=webp&s=de5568e69a4bec51701ef90a86f53464ea06716d

Another guy on that same stack exchange says need to be told it is ideal gas first: https://physics.stackexchange.com/questions/113586/why-change-in-internal-energy-is-zero-in-isothermal-process

https://preview.redd.it/tw17tefoczt41.png?width=1329&format=png&auto=webp&s=0fc4d1f1750ed0bf32dfc908aff298ff889e9e80

On the MCAT, is it safe to say "oh ok isothermal -> delta U =0" or do we need to be told it is ideal gas first to use this?

This has been confusing me for an hour...

If the gas container is heated with a lighter - so Heat (Q) increases for the particles - that heat instantly moves out of the system to make the Temperatures outside and inside the system the same (otherwise it's not isothermal lol). So in that case, how is the Heat (Q) being converted to Work (W) (aka increasing Volume) if the Heat (Q) is just being transferred outward?

sry, i am dumb but plz help me ty <3

I thought work and heat, being path functions are inexact differentials. Why can we integrate for a reversible isothermal process when Macquarrie's physical chemistry says that we can't. What am I missing conceptually here?

Hello everyone,

Picture following situation: in a transient process, air flows into a rigid tank and then rapidly cools down to 300K.

My professor told me that because of the heat transfer being fast, the process can be regarded as isothermal, which makes the temperature of the air, that is already in tank, before the filling process also 300K.

My question is: Can a quick heat exchange always be regarded as isothermal or is there something I overlook?

I got stuck solving the problem: https://imgur.com/LUSB762

Help.

Thanks!

Say that during an adiabatic process, the pressure changes from 𝑝0 to 𝑎∗𝑝0, and let the heat capacities ratio be denoted by b ; using the basic adiabatic relations for an ideal gas, we can find that the temperature changes from 𝑇0 to 𝑎^(1−1/𝑏). As the degrees of freedom of a molecule in the gas tend to infinity, b will tend to 1, which implies that for high degrees of freedom there is little to no temperature change during an adiabatic process i.e. it becomes isothermal. Why is this?

My intuition tells me that high D.O.F. implies more places to store energy, requiring more energy for a given V,P,& T; hence an adiabatic process w/ high D.O.F. will have so much stored energy that the temperature change caused by moving from 𝑝0 to 𝑎∗𝑝0 will be negligible. Is this the right idea?

On a different note, how valid is the ideal gas law when it comes to gases with high D.O.F ? What gases have the highest D.O.F ?

Hi! I'm doing a homework for my math class about application of integral calculus and I chose isothermal processes.

Iunderstand that the work is calculated by getting the definite integral of the curve in pV graph.

However, I wanna ask what are the practical applications of such where they have to compute the amount of work done?

Thanks in advance!

I don't get the difference between adiabatic and isothermal processes. Specifically, for adiabatic, how could there be a temperature change if heat cannot be transferred?

I need to find the work done by the gas in a complete cycle, and to do that i need to find the temperature at point B and C. Since it is a isothermal process the temperature is the same at both. When i use PV=nRT i get 12000 K and when i use P1/T1=P2/P2 i get 1200K. Both seem wrong what would i use to get the temperature? https://imgur.com/zYD3Yjd

Is it possible to have isobaric, isothermal and isomeric processes in one system?

I'm stuck. I know the isothermal process shouldn't be isenthalpic, but enthalpy is defined by h = u + pv, right? In an ideal gas, if the temperature is fixed, u should remain constant, and pv also remains constant...right?

I get the feeling I'm missing something obvious. Help?

Hi All

The basic assumption when pipe flow is discussed that the process is isothermal. Whether those are Poiseuille flow for the laminar case, or viscous Bernoulli relations - we assume that the temperature stays the same

But this is not the case always. Sometimes the flow would heat. So, my question is - are there relatively easy ways to try and estimate it? Because right now the only way that I can think of is to calculated local viscous dissipation, run an integral on it, assume adiabatic walls and get it from there.

So, I see that microcanonical has constant and known energy, volume and number of moles and is therefore isolated energetically and it cannot let particles out or in. But canonical has constant known temperature in place of energy and therefore isothermal, which is not the same as adiabatic and energy is allowed to flow. But in isothermal processes the internal energy is the same and therefore the average kinetic energy of the molecules stays the same and if there is net heat flow, then there has to be work done, but the volume is constant, so work cannot be done, so how can there be net heat flow at all? If we know the volume, temperature and number of moles, we should also know the energy of the system. I am certain I am missing something, so if anyone can help, I will be grateful.

What Is The Relationship Between Entropy And Black-holes?

Black-hole entropy is related to its surface-area ‘A’ in accordance with the Berkenstein-Hawking relationship (below). The surface of a black-hole consists of units of entropy equal to the Planck-length, squared, ‘L²’, in accordance with the Hawking equation...

S(bh) = ¼(A/L²), where ‘A’ is the area of a black-hole (bh).

According to an extension of this view, photons absorbed by a black-hole join the EH-boundary envelope and serve as work to increase its size; photon energy is presumed to represent entropy; as well as work to expand the black-hole.

Or one may, perhaps, regard this work as an isothermal process with zero entropy as a consequence of the work done. Entropy (photon-energy), in this case, would then represent the internal energy of a black-hole. The classic definition of entropy as a by-product of work (the Carnot cycle for example) may not apply to black-hole physics.

NOTE: The black-hole described in this post is a departure from current BH physics; however, it accommodates accepted principles of math, physics and logic in addressing untenable conclusions offered by current BH concepts.