Adiabatic Process Puns

Question:

Suppose 2.00 mol of a monatomic ideal gas (cV = 3/2R) is expanded adiabatically and reversibly from a temperature T = 300 K, where the volume of the system is 20.0 L, to a volume of 60.0 L. Calculate the final temperature of the gas, the work done on the gas, and the energy and enthalpy changes.

I calculated the final temperature by using the following equation:

T1V1^(cp/cv)-1 = T2V2^(cp/cv)-1

In which I got Tfinal = 144 K (I checked the answer key and this is correct)

However I am stuck on finding deltaH or w = deltaU (since the system is an adiabatic process, q=0)

I tried using various methods but none were resulting in the correct answer (the answer key reports that w/deltaU = 23.89 kJ and deltaH = 26.48 kJ)

For reference, these are some methods I tried to use:

> deltaU = n*cv*deltaT

>deltaH = n*cp*deltaT

>w = -n*R*T*ln(V2/V1) (Yes, I know this relationship only applies to isothermal processes, but I exhausted all the options that I could think of, so I gave it a shot)

Hi! I’m currently working on a project based on the automarizarion of evaporative air coolers. I’m trying to create a mathematical model that can help me interpret a psychrometric chart/ get the results I’d get from one using equations. Has anybody done this before? Or does anyone have any tips? Thanks!

So there are 2 processes for 1kg of Water. 1st is adiabatic and second is isobar. Everything that is given:

p1= 0.1MPa , p2 = p3 = 1MPa, T1 = 100C and T3 = 400C.. how do I calculate T2 without calculating the Isentropic expontent Kappa (or K whatever you like). I can see the Volume V3 from some given Table but I would need V2 how do I get it?

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air is ideal (cv=716 J/kg*K and R=287 J/kg*K) and water is incompressible (c=4187 J/kg*K) and rho=1000 kg/m^3). What kind of calculations can be done to prove that this is an adiabatic process?

https://preview.redd.it/1lecan4myu361.png?width=1094&format=png&auto=webp&s=42f5707ed0eef1ff990922e7bd77fe47a6524734

So what I currently know is that the adiabatic relation is pv^gamma = constant.

Applying this between 2 states the following is obtained: p1 v1 ^ gamma = p2 v2 ^ gamma

So, if there is say a closed vessel, the total volume will be constant: v1 = v2

Putting this relationship into the adiabatic relationship the volume ^ gamma would cancel leaving p1 = p2. This means that the pressure is also constant?

If this is true, I got this question where there is a closed vessel and there is some methane and air. It’s ignited with a spark to cause combustion but the process is assumed to be adiabatic. So here the volume is constant, then from above the pressure is also constant therefore if we apply the ideal gas equation between the 2 states (state where there are reactants and another where only products), the number of moles remains the same therefore the temperature would also remain the same (according to the derivation above if volume and pressure are constant). But this can’t be.

I’m not sure where I’m going wrong? Can anyone please help me, it would be greatly appreciated! Thank you.

When we say " isentropic process" ,our mind goes to think internal reversible and adiabatic process. It is right but they aren't two faces for one coin. However the relationship is every reversible adiabatic process is necessarily isentropic process. But not every isentropic process is necessarily reversible adiabatic process. That explains to fix entropy values, we have many state First: reversible adiabatic .reversible to make (entropy generation equals zero) What means when I reject heat and absorb it again entropy values don't change. And adiabatic to not making losses or gaining of entropy. Second :To make entropy values fixed we should make balance between the increase and decrease of entropy So the source of increase is entropy generation or gaining heat to system. And the source of decrease is lossing heat from the system. So we find that we use the way of lossing heat to decrease entropy values by the same amount we increase entropy by entropy generation. Do you agree with me? Am I right? Can you find another state for Isentropic process? If I had a grammatical mistakes because I am not native speaker

I am studying thermodynamics for ChemE, and I am confused by a CHEGG response I have encountered when calculating dH, dU, dQ, and dW for a process path.

Fundamentally, for an adiabatic process path, I have always believed that ONLY dQ is zero. The defined system temperature may change (dT), but there is no heat exchange between the system and the Environment regardless of the change in T (dQ=O).

I have run into a chegg response that says there is a non-zero dQ for this adiabatic process path which is = dW. They also say that dU=O and dH=O. The explanation they give is that dT is always 0 for adiabatic process paths, which obviously contradicts what I have believed all through my first ChemE semester. I’m lost because it seems so fundamentally wrong, but it has like 18 thumbs up and no thumbs down on the answer rating.

In summary, do adiabatic process paths mean dQ=0 in all circumstances, or does it mean dQ and dT= 0 under all circumstances?

Am I missing something?

If anyone has chegg, it is chapter 3 problem 21b in “introduction to chemical engineering thermodynamics 8th edition by smith and van ness”.

My answers for a) and b) are in the imgur link (as well as the question). the next question wants me to find the work done in the process C, which is adiabatic. in my notes for adiabatic processes I have that Q = 0 and that E(internal) = W. But then I looked at the answers our teacher gave us and he wrote :

"For cycle E(int) = 0

-> W = -(-3286) = 3286 J"

But I'm confused why he's done W = negative of the value I got for the heat transferred isovolumetric process (which I calculated in b)???

https://imgur.com/a/M14PP60

I'm going to attempt to teach you everything I know.. as long as you guys are willing to learn! I taught my brother a lot of stuff after I got out of the Marine Corps and he got the hang of it pretty quickly.

Alright, so here's the break it down barney style of the adiabatic process.. I promise it's not difficult at all

Adiabatic process

I want to point out that the adiabatic process works both ways. This is literally how clouds form. It also has a huge part of hurricane formation.. they need an environment that can cool adiabatically, just like a regular thunderstorm, but hurricanes need a mesoscale sized area of it for their entire lifespan. However, after the parcel has reached saturation, it will cool slower due to having to condense out moisture while releasing latent heat

ILX SKEW T, from 1200z today

Skew-ts are a bit odd to grasp at first, but notice the constant lines of temperature are skewed to the right. They are the orange dashed ones, the right handed blue one is 0C and the left one is -20C as shown on the bottom of the chart.

As your red line turns sharply to the right to the point where it isn't parallel to the temperature lines, that shows the temperature is rising, to the left is falling. Same with the green line with the skew t. The difference in distance from the red one shows how much moisture (or lack of distance I should say) you have.

Notice how at the end of the brown arrow the green line turns to the left and the red line sharply turns to the right.

Now, the troposphere (where we live, bottom layer), the temperature naturally falls with height. In the stratosphere the temperatures naturally rise with height.

When you have a high amplitude wave in your jet stream, it will cause the stratosphere sink intensely, which causes the sharper inversion at the tropopause (where the inversion happens) and also causes the tropopause to fall with height. The dry air from the stratosphere then sinks into the troposphere, creating a strong high pressure sinking in from the north.

Water Vapor Satellite shot from today

This is typically a winter event. Any questions?

P.S. I literally just made up numbers from the box. I don't wanna do math right now.

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 ?

The title basically.

I know the classic representation in pV, but I can't seem to find the other two

Hey everyone. I’m a meteorologist in the U.S. Marine Corps and I’ve been forecasting weather for about 3 years now. I’m an enlisted Marine, I do not have a college degree so most of the knowledge I have is from my senior Marines and me learning as much as I can about atmospheric thermodynamics. For the Marines out reading this... I’m an east coast Marine. Yut.

I thought I’d drop by and explain to /r/weather what the adiabatic process is and how it differs from the diabatic process. First off I’ll put it in simple terms, air that falls adiabatically will warm and air that rises adiabatically will cool. So keeping that in mind, the diabatic process should be the opposite. The diabatic process is a change of temperature in a parcel that doesn’t have anything to do with the way it moves up and down. The adiabatic process has to do with the ideal gas law; PV=nRT. In the adiabatic process, the parcel of air does not have any heat exchange with the surroundings. So imagine 5 little dots floating around in a circle, as this circle rises the pressure on it is less, right? So the little dots hit the side of the circle less which makes the temperature of the air parcel drop. Same vice versa with the circle with the same amount of dots sinking, because the volume of the air parcel dropped, the temperature of it will rise due to the dots hitting the circle more. Okay that’s about as Barney style as I can get it.

The diabatic process has to do with entropy, which means that energy will flow to where there is less of it to reach an equilibrium. So imagine this… you take a cooler of ice from the case of beer you just got done drinking and dumped it on your kitchen floor of your house because your A/C is out in the middle of summer in the southeastern United States. You’re extremely drunk and hot… but you notice the air around the ice is colder, why? Well this is the diabatic process. As the ice is melting it needs to get the energy from somewhere to melt it, so it gets it from the surrounding air! So when you wake up the next morning hungover as hell, you notice the puddle of water in your kitchen floor is the same temperature as your air temperature.

So I bet everyone is wondering how this has to do with weather. I like to use the southeastern United States because that’s where I’m from and I’m still stationed in the region. This will help everyone that lives on the beach, does afternoon thunderstorms ring a bell? It’s hot and muggy in the middle of July on the east c

I believe that there is an error here. Here's the question: "An ideal monatomic gas at p1=3E5 N/m^2, V1=0.06 m^3, and T=300 K expands adiabatically to p2=2E5 N/m^2, V2=0.08 5m^3, and then isothermally to V3=0.1 m^3. What is the final T3 and p3?"

For the first part, I have this: P1=3E5 P2=2E5 V1=0.06 V2=0.085 T1=300 T2 -----> We must use the adiabatic equation below.

T1V1^0.4 = T2V2^0.4 ---> T2 = 261 K

The second part of the problem is now an isothermic problem, and we can use the ideal gas equation.

P2 = 2E5 P3 = ? V2= 0.085 V3 = 0.100 T2 = 261K T3 = 263

P3 = P2*V2/V3 ---> P3 = 1.7E5 N/m^2

However, the book says this:

"First calculate the temperature at the end of the adiabatic expansion using the ideal gas law:"

"P1V1/T1 = P2V2/T2 ---> T2 = 140K"

NOTE: I HAVE NO IDEA HOW THEY GOT THIS. THIS LOOKS LIKE AN ERROR AT TWO LEVELS. (1) BECAUSE YOU CAN'T USE THE IDEAL GAS LAW HERE SINCE IT'S ADIABATIC. ALSO, (2) T2 = T1P2V2/P1*V1 = 283K .

I have a bottle filled with an ideal gas in a vacuum. On the top of the bottle a long pipe is attached. A perfectly fitting/sealing bullet is released in this pipe.

The bullet is released. As it falls it causes an adiabatic compression of the gas. After a time 't' the bullet is pushed back to its initial position. Write 't' in function of : Area, volume, pressure, mass of bullet, adiabatic constant gamma.

EDIT: The bullet is not shot or something, just released in a freefall. / Mild calculus and/or differential eq

This is just for fun, not homework. So have a go and see how you stack up against the community!

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