would be good to add that the reason an adiabatic process is steeper than the iso-thermal process is due to the decrease or increase in gas particle velocity (which affects pressure) caused by the decrease or increase in temperature.
Hi, please take note that there is a mistake at 12:11. both isothermal and adiabatic curves should be single arrow. during compression process, the work done of isothermal process is less than the work done of adiabatic process, so the adiabatic curve should above isothermal curve in the revers way.
No, that's not a mistake. He used two arrows to symbolize compression and expansion in both isothermal and adiabatic processes. At least, that's what I think.
Hi quick version of getting work in an Isothermal process when given the change in V ,number of moles and the temperature is using the equation W=nRTln[vf/vi];
Do you think that internal energy remain same and work done by heat exchange in isothermal process, then think again. Heat exchange takes place when there is difference in temperature but in isothermal, temperature is same. Work done is at cost of internal energy, change in pressure.
Actually, this depends completely on your perception. Conventionally, when work is done ON the system, it is taken to be positive, because, the system gains energy as, U=Q+W. On the other hand, when work is done BY the system, it is taken to be negative, as energy is lost by the system, giving rise to the relationship, U= Q- W. However, if you take the first law of thermodynamics as: U= Q-W, then, your assumptions is correct. If work done on the system is taken as negative, then, U=Q-(-W) i.e. U= Q+W. This completely depends on your perception, and honestly, convention does not matter as long as your conception is sound.
I am stuck with first law of thermodynamics because of its mathematical expression, some are using like above expression which is u= Q+w and remaining are using U= Q- W
It's just some difference we have in subjects ...like the plus one is for phy and negative for Chem.....it's just differs on tge respective of where you are seeing or basically the point of view :)
harly moon isothermal means there is no change in temperature, hence no change in internal energy, U. dU is therefore 0, yet heat might still flow, so dQ isn't zero.
Hey! Just wanted to point out something. During your isothermal description you rearranged the formula wrong, should be U=Q-W. Great vid though! Helped out alot :)
U=Q+W is true because he said W= work done "on" gas... For U=Q-W, W= work done "by" the gas... its just a matter of how u want W to be represented by...😉
isn't it that the formula for 1st law of thermo dynamics is Q= change in U + WORK? how come that change in U is now equal to heat plus work. ? it should be Q-W ? please correct me if I'm mistaken. Thank you.
The equation is U= Q + W and the W can be positive or negative depending on what the gas does. If the gas did the work, then work is negative. If the work was done on the gas then the Work is positive. Hope that helps
but when we compressed the tank W is negative, the equation is delta U= Q-W ( W is in negative since work is done "on system") thus it becomes positive and not delta U= Q+W and since heat is flowing out so Q is negative , just wanted to clarify on delta U= Q-W , please reply
Husam because at constant volume, change in volume is 0. the p🔺v term goes to 0 leaving u=q . so at constant volume, no matter what work you do, it wont affect your system.
Ohhhhhhh, for the life of me I was super confused about it, but yeah if the final and initial volume stays the same then that will be 0. Now it sounds so simple!
Good question. The adiabatic curve is given by the relationship of P*V^k = constant, where k is the ratio of specific heat capacities, defined as k=cp/cv. Since the constant pressure specific heat is always greater than the constant volume specific heat, due to the gas expanding as it is heated under constant pressure, the value of k will always be greater than 1. Isothermal is given by P*V = constant (straight from the ideal gas law), while P*V^k = constant generates a steeper curve that governs the adiabatic reversible process. Why does P*V^k govern adiabatic processes? Here's the derivation. hyperphysics.phy-astr.gsu.edu/hbase/thermo/adiabc.html#c1
The gamma ratio (some books call it k), is the ratio of specific heat capacities, that governs the reversible adiabatic process for an ideal gas. It is defined as gamma = cp/cv, as the ratio of constant pressure specific heat capacity, to constant volume specific heat capacity. It generally is a function of the shape of the gas molecule, such that 2-atom gasses like air and its components have gamma = 1.4, and single atom gasses like Argon, have gamma = 1.67. It has to do with the degrees of freedom in the gas molecule, that act as modes of energy storage.
Holy sh** this was so well explained that even my cat would understand it!
would be good to add that the reason an adiabatic process is steeper than the iso-thermal process is due to the decrease or increase in gas particle velocity (which affects pressure) caused by the decrease or increase in temperature.
Was literally gonna ask this, my g
thanks for the info, i had just begun wondering why pressure ended up higher for adiabatic expansion
You saved my life.
7:40 You are a) extremely good at your explanations and b)very enthusiastic/funny at times
I wish there were more physics derivation videos from you specifically
Hi, please take note that there is a mistake at 12:11. both isothermal and adiabatic curves should be single arrow. during compression process, the work done of isothermal process is less than the work done of adiabatic process, so the adiabatic curve should above isothermal curve in the revers way.
No, that's not a mistake. He used two arrows to symbolize compression and expansion in both isothermal and adiabatic processes. At least, that's what I think.
Nice video which I understand this PV diagram for the first time thanks a lot good one 👍😍
ihave my thermo exam today thanks
Thank you for explaining so perfectly!
this is very helpful when your teachers method of teaching is reading word for word off a PowerPoint and jarringly bouncing between slides constantly.
What a tremendous !!!
video best explanation ever!!!😇
Why does this not have more views
I am so grateful for finding this video haha. Thanks!
Thanks for explaining these so well!
Omg..... U guys are so good at teaching 👍👍👍👍🙏🙏🙏🙏
soooo useful, I love this channel
All other vids kept confusing me.TYSM
This is a really good video. Very clear and easy to understand.
Hi quick version of getting work in an Isothermal process when given the change in V ,number of moles and the temperature is using the equation W=nRTln[vf/vi];
8:20 you mean deltaQ=U+W
Awesome video, just what I was looking for!!
So easy to understand
Do you think that internal energy remain same and work done by heat exchange in isothermal process, then think again.
Heat exchange takes place when there is difference in temperature but in isothermal, temperature is same. Work done is at cost of internal energy, change in pressure.
Do you now somthing ...u are awesome....thank sooo much
Thank you very much...!!!!!
I shouldve watched your vid before I spent 3 hrs of my life trying to figure out what these three are.
Thank you good sir. 👍🤘
Well done ...
Good explanation
Well Done - thank you for explaining thermo and making it easy to apprehend. Where were you when I was in undergrad?!
Is apprehend means the same thing as i think it means?
This was a good video but I think the equation for (delta U=Q+W) is wrong I think it's suppose to be minus
isn't it delta U = -W since work is done onto the system cause work isn't done BY the system?
Actually, this depends completely on your perception. Conventionally, when work is done ON the system, it is taken to be positive, because, the system gains energy as, U=Q+W. On the other hand, when work is done BY the system, it is taken to be negative, as energy is lost by the system, giving rise to the relationship, U= Q- W.
However, if you take the first law of thermodynamics as: U= Q-W, then, your assumptions is correct. If work done on the system is taken as negative, then, U=Q-(-W) i.e. U= Q+W. This completely depends on your perception, and honestly, convention does not matter as long as your conception is sound.
thank you!
excellent
thank you
Thank you thank you thank you very much
Does the GHE effect earth's temperature, or is earth's mean temperature of 15 C controlled by the atmosphere's mass and pressure from gravity?
brother. what is the name of this Software. you are writing like pen by mouse Amazing.
Thank you for this video. My professor thought it was cool to use a shitty graph and an unintuitive explanation to help us understand this.
nice!
Hey can u please tell 🙏🙏🙏. Where do u write this I mean what is the software.
At 8:22, did you mean that U = Q - W?
Although i didnt ask the question, thank you for your reply. I was getting proper confused, but now i understand.
I am stuck with first law of thermodynamics because of its mathematical expression, some are using like above expression which is u= Q+w and remaining are using U= Q- W
They put negative on it because of W=-P(Vf-Vi) iguess?.
It's just some difference we have in subjects ...like the plus one is for phy and negative for Chem.....it's just differs on tge respective of where you are seeing or basically the point of view :)
I can't believe these videos are uploaded 7 years ago
thank u for help me
This is a isothermal process. So the eqution should be like
dQ=dW(where du=0)
Sajim Ahamed no I think it should be W=-∆U where (Q=0)
harly moon i need sometime to think about it.
harly moon isothermal means there is no change in temperature, hence no change in internal energy, U. dU is therefore 0, yet heat might still flow, so dQ isn't zero.
For some reason, I thought the isothermal & adiabatic curves are switched, meaning tha adiabatic requires more work. I guess I was wrong.
Hey! Just wanted to point out something. During your isothermal description you rearranged the formula wrong, should be U=Q-W. Great vid though! Helped out alot :)
ohh thanks..im confuse for a little bit because of that while watching this
U=Q+W is true because he said W= work done "on" gas...
For U=Q-W, W= work done "by" the gas...
its just a matter of how u want W to be represented by...😉
helpfull
Thanks a lot!
isn't it that the formula for 1st law of thermo dynamics is Q= change in U + WORK?
how come that change in U is now equal to heat plus work. ? it should be Q-W ? please correct me if I'm mistaken. Thank you.
The equation is U= Q + W and the W can be positive or negative depending on what the gas does. If the gas did the work, then work is negative. If the work was done on the gas then the Work is positive. Hope that helps
I thought the first law of thermodynamics was delta U = Q - W
How is pressure constant in Isochoric process?
It isn't. It is volume that is constant in an isochoric process.
water temp outside container increase or not
yes, but very little, unless you have a small amount of it
but when we compressed the tank W is negative, the equation is delta U= Q-W ( W is in negative since work is done "on system") thus it becomes positive and not delta U= Q+W and since heat is flowing out so Q is negative , just wanted to clarify on delta U= Q-W , please reply
Why is that at a constant volume no work can be done but at a constant pressure work can be done?
Husam because at constant volume, change in volume is 0. the p🔺v term goes to 0 leaving u=q . so at constant volume, no matter what work you do, it wont affect your system.
Ohhhhhhh, for the life of me I was super confused about it, but yeah if the final and initial volume stays the same then that will be 0. Now it sounds so simple!
But first law of thermodynamic States that dQ=W+dU , so how dU=W in an adiabatic process
Do you mean why is it not dU=-W? Because I do not understand why it is not a negative W value.
Because in an adiabatic process, heat change is 0, hence dq=0.
Therefore 0=dU+W
W=-dU
wll explained
why adiabatic curve is more steeper than isothermal curve ???
Good question. The adiabatic curve is given by the relationship of P*V^k = constant, where k is the ratio of specific heat capacities, defined as k=cp/cv. Since the constant pressure specific heat is always greater than the constant volume specific heat, due to the gas expanding as it is heated under constant pressure, the value of k will always be greater than 1. Isothermal is given by P*V = constant (straight from the ideal gas law), while P*V^k = constant generates a steeper curve that governs the adiabatic reversible process.
Why does P*V^k govern adiabatic processes? Here's the derivation.
hyperphysics.phy-astr.gsu.edu/hbase/thermo/adiabc.html#c1
ohhhhhhhhh my god thank u!
Is the Q component heat transfer?
+The man himself yes, thermal energy
what about gamma ratio?
The gamma ratio (some books call it k), is the ratio of specific heat capacities, that governs the reversible adiabatic process for an ideal gas. It is defined as gamma = cp/cv, as the ratio of constant pressure specific heat capacity, to constant volume specific heat capacity. It generally is a function of the shape of the gas molecule, such that 2-atom gasses like air and its components have gamma = 1.4, and single atom gasses like Argon, have gamma = 1.67. It has to do with the degrees of freedom in the gas molecule, that act as modes of energy storage.
Damn good explanation. I studied engineering at Cornell. Trying to coach my son’s Science Olympiad Thermodynamics team.
👌👌👌
❤
Holy Jesus if only i searched and watched these in 11th grade 😭😭
Edit: wasted money on shi*ty tutions
Holy fuck, thanks man, truly from the ravine of my heart!!!!
LOOKIT. NOW LOOKIT. LOOKIT
du=dQ-dW.
Swarna Biswas U
Swarna Biswas That sign depends on what nomenclature you are using for work & heat with regard to the system.
sangram tayade +yeah.thanks.
lol kaplan had 2 words on this
rest of theexplanation was very good
why adiabatic graph is more steeper?????????
It is because of the instant action required to do it. NOT SURE BUT!
Here's the derivation of the adiabatic condition:
hyperphysics.phy-astr.gsu.edu/hbase/thermo/adiabc.html#c1
Thank you very much...!!!!!
Thank u
V good