Thank you so much for everything you have done Mr. van Biezen. I wouldn't be able to be where I am today if it wasn't for your wonderful teaching. Thank you so much!
In my textbook, the process where volume doesn't change is also called isochoric. Sorry sir I used the image of one of your videos because I loved the pulley problem.
Does a HE cycle with a Lower entropy change compared to a one with a higher entropy change always mean one with a lower entropy change is more efficient HE cycle? Thanks
Hello ! I have a little question: Why would the temperature decrease if we decrease the volume ? Wasn't it that when a gas is compressed, the temperature increases ?
The first graph on the left shows that the temperature increases when the volume increases or when the pressure increases (as the volume remains the same). If the volume decreases (at the same pressure) the temperature will decrease.
@@MichelvanBiezen So when the pressure is constant, the kinetic energy of the molecules will decrease( when we decrease the volume), therefore the temperature will decrease ?
Do you have a special sequence to watch the videos? I want to learn the basics of thermodynamics, heat transfer and fluids. Let me know. I will appretiate it.
Fransisco. If you go to any search engine (Yahoo, Google, Chrome, etc.) and you search for "Michel van Biezen TH-cam" and then click on "Michel van Biezen - TH-cam" you get to the channel main page. There you'll find all of the main play lists as well as a summary of all the playlists. There is one on Thermodynamics and One on Fluids, as well as many others.
Dr.Michel we said in this video that direction of temperature will be positive sign as we go to the right or up. But here volume is going from 15 L to 2L like that it will go to left and temperature has to have a negative sign. Could yo clarify this point? Thanks alot.
Yes you are correct. The problem was set up in a "general" fashion, and the work calculated was negative which indicates that work was done ON the gas and that the volume of the gas was reduced, which means the direction is from right to left.
Suppose you did this problem in two steps, since U is a state function. Suppose the first step were heating at constant volume and the second step were isothermal expansion. You should get the same answer, right? But using the two-step process I get delta U = -800 J for the first step because P delta V = 0; for the second step I get delta U = 0 because U only depends on temperature. What am I doing wrong in my two-step calculation, please?
Whoops, I see where I went wrong. In my first step (constant volume) I used U = q - w assuming that q was -800 J. But I should have calculated the new constant-volume q, because q was only -800 J in the isobaric situation. To calculate the q for a constant V step, I should have found the delta T using q = n*Cp*delta T first (because that's what really happened in the experiment), then then used that delta T to find the constant-V version of q, from q = n*Cv*delta T.
The thing about isobaric never gets me. I cant quite imagine a system of isobaric system where compression will only result in change in volume without concmittant change in pressure. anyway , interstingly it seems theorticallly relevent here. I would imagine a very thermoelastic system where where the work done on the system PdV would result in liberation of heat, and change in its internal temperature, without having to change its pressure....
Michel, In my textbook it states that Q=Cm(Tf-Ti) where m is mass and C is specific heat or heat capacity per unit mass. What the difference between the books equation and the one you are teaching? So you have moles instead of mass.
Abdel, The equation Q=cm(Tf-Ti) is used for solids and liquids. (m = mass and c = # of joules / (kg * K) The equation Q = cn(Tf-Ti) is used for gases. (n = the number of moles) and c depends on the type of gas (monatomic, diatomic, triatomic, etc.) and what type of process (constant volume or constant pressure). Sometimes this is confused with the equation delta U = nc(Tf-Ti) which is used for the change in internal energy where c = cv (the same constant used for the constant volume process).
Given Q is negative, then (delta T) is negative from Q = n(c_p)(delta T). Given (delta U) is positive, then (delta T) is positive from (delta U) = n(c_v)(delta T). Did the temperature increase or decrease?
+Steven Harding It all depends on the definition of Q (unfortunately, there are 2 definitions) I use the following definition: Q is positive when heat is added TO the gas. In this example, Q must therefore be positive. Next, in any process where one moves to the right or up on the PV diagram, the change in temperature will therefore be positive. Since the change in temperature is positive, delta U must also be positive.
When I plot the answer in the PV diagram. The isobaric solution make no sense. According to the PV diagram final temperature should be less than the final but according to the calculation since we got positive q the temperature is increasing. Can you tell me where I am going wrong please. Thank you.
dnonod1 In the PV diagram. When the new state is further to the right (greater volume), the temperature will be greater When the new state is further up (greater pressure) the temperature will be greater. Isotherms have a curve that looks like a 1/x curve. In an isobaric process, if the volume increases the temperature will be higher In an isobaric process, if the volume decreases the temperature will be lower.
The problem statement is incorrect. You have over-defined the isobaric process and the parameters are in conflict. The next video in this playlist states the problem correctly (without the prescribed heat exchange) and solves it correctly. In principle, if you didn't assume an isobaric process here, the problem wouldn't have been ill-defined.
In summary its actually that the actual energy applied to the system is by compression that is Pdv = 1976Joules. From that 1176 Joues was given to internal energy of system and 800 Joule being liberated as heat...practaically quite unralistic since compression would usually result in change in pressure, but theoritically thats how i imagined it.
Thank you so much for everything you have done Mr. van Biezen. I wouldn't be able to be where I am today if it wasn't for your wonderful teaching. Thank you so much!
You are very welcome. Thanks for sharing.
Where are you today?
These videos are isobarrific! 👍🍫
Thank you.
You, Sir, are awesome.
Sir, you are an amazing teacher
In my textbook, the process where volume doesn't change is also called isochoric. Sorry sir I used the image of one of your videos because I loved the pulley problem.
We will consider it as a highest form of flattery.
@@MichelvanBiezen Haha I'm quite relieved that you're mad at me for "stealing" your image. Thank you and nice videos on all of the subjects!
Does a HE cycle with a Lower entropy change compared to a one with a higher entropy change always mean one with a lower entropy change is more efficient HE cycle? Thanks
Hello ! I have a little question: Why would the temperature decrease if we decrease the volume ? Wasn't it that when a gas is compressed, the temperature increases ?
The first graph on the left shows that the temperature increases when the volume increases or when the pressure increases (as the volume remains the same). If the volume decreases (at the same pressure) the temperature will decrease.
@@MichelvanBiezen So when the pressure is constant, the kinetic energy of the molecules will decrease( when we decrease the volume), therefore the temperature will decrease ?
It is more like this: If you lower the temperature of the gas and keep the pressure of the gas constant, the volume will decrease.
Now i get it ! :D Thank you, Mr Biezen ! Continue making videos for us, i am sure we are all thankful for your support !
Do you have a special sequence to watch the videos? I want to learn the basics of thermodynamics, heat transfer and fluids. Let me know. I will appretiate it.
Fransisco. If you go to any search engine (Yahoo, Google, Chrome, etc.) and you search for "Michel van Biezen TH-cam" and then click on "Michel van Biezen - TH-cam" you get to the channel main page. There you'll find all of the main play lists as well as a summary of all the playlists. There is one on Thermodynamics and One on Fluids, as well as many others.
Dr.Michel we said in this video that direction of temperature will be positive sign as we go to the right or up. But here volume is going from 15 L to 2L like that it will go to left and temperature has to have a negative sign. Could yo clarify this point? Thanks alot.
Yes you are correct. The problem was set up in a "general" fashion, and the work calculated was negative which indicates that work was done ON the gas and that the volume of the gas was reduced, which means the direction is from right to left.
Sir, delta U has to be negative. The internal energy must decrease during isobaric compression. T2
I think he calculated work wrong, 1 litre = 0.001m3, his conversion is wrong
HelloHow did they derive the equation v2/v1 = T2/T1
That comes from the ideal gas equation: PV = nRT If P (Pressure) is constant then V/T = nR/P = constant and V1/T1 = V2/T2
In this video you should find the volym, its a isobaric stat???
Suppose you did this problem in two steps, since U is a state function. Suppose the first step were heating at constant volume and the second step were isothermal expansion. You should get the same answer, right? But using the two-step process I get delta U = -800 J for the first step because P delta V = 0; for the second step I get delta U = 0 because U only depends on temperature. What am I doing wrong in my two-step calculation, please?
Whoops, I see where I went wrong. In my first step (constant volume) I used U = q - w assuming that q was -800 J. But I should have calculated the new constant-volume q, because q was only -800 J in the isobaric situation. To calculate the q for a constant V step, I should have found the delta T using q = n*Cp*delta T first (because that's what really happened in the experiment), then then used that delta T to find the constant-V version of q, from q = n*Cv*delta T.
The thing about isobaric never gets me. I cant quite imagine a system of isobaric system where compression will only result in change in volume without concmittant change in pressure. anyway , interstingly it seems theorticallly relevent here. I would imagine a very thermoelastic system where where the work done on the system PdV would result in liberation of heat, and change in its internal temperature, without having to change its pressure....
there is also heat transfer
Michel,
In my textbook it states that Q=Cm(Tf-Ti) where m is mass and C is specific heat or heat capacity per unit mass. What the difference between the books equation and the one you are teaching? So you have moles instead of mass.
Abdel,
The equation Q=cm(Tf-Ti) is used for solids and liquids. (m = mass and c = # of joules / (kg * K)
The equation Q = cn(Tf-Ti) is used for gases. (n = the number of moles) and
c depends on the type of gas (monatomic, diatomic, triatomic, etc.) and what type of process (constant volume or constant pressure).
Sometimes this is confused with the equation delta U = nc(Tf-Ti) which is used for the change in internal energy where c = cv (the same constant used for the constant volume process).
Michel van Biezen Thank you Michel! I found the equation in this video in my textbook, it was in the following chapter.
Given Q is negative, then (delta T) is negative from Q = n(c_p)(delta T). Given (delta U) is positive, then (delta T) is positive from (delta U) = n(c_v)(delta T). Did the temperature increase or decrease?
+Steven Harding Unless the original problem was to find (P_2), thereby showing that this state change is not isobaric
+Steven Harding It all depends on the definition of Q (unfortunately, there are 2 definitions) I use the following definition: Q is positive when heat is added TO the gas. In this example, Q must therefore be positive. Next, in any process where one moves to the right or up on the PV diagram, the change in temperature will therefore be positive. Since the change in temperature is positive, delta U must also be positive.
When I plot the answer in the PV diagram. The isobaric solution make no sense. According to the PV diagram final temperature should be less than the final but according to the calculation since we got positive q the temperature is increasing. Can you tell me where I am going wrong please. Thank you.
dnonod1
In the PV diagram. When the new state is further to the right (greater volume), the temperature will be greater
When the new state is further up (greater pressure) the temperature will be greater.
Isotherms have a curve that looks like a 1/x curve.
In an isobaric process, if the volume increases the temperature will be higher
In an isobaric process, if the volume decreases the temperature will be lower.
The problem statement is incorrect. You have over-defined the isobaric process and the parameters are in conflict. The next video in this playlist states the problem correctly (without the prescribed heat exchange) and solves it correctly. In principle, if you didn't assume an isobaric process here, the problem wouldn't have been ill-defined.
In summary its actually that the actual energy applied to the system is by compression that is Pdv = 1976Joules. From that 1176 Joues was given to internal energy of system and 800 Joule being liberated as heat...practaically quite unralistic since compression would usually result in change in pressure, but theoritically thats how i imagined it.
why delta u=ncvt?
In an isovolumetric process, W = 0 and the firs law of thermodynamics states: delta U = Q - W Since W = 0, delta U = Q = nCv delta T
Michel van Biezen thank you sir,now its clear...
Michel van Biezen its a formula for internal energy,can i use this at any conditin?
Yes that formula for the internal energy ALWAYS correct.
thank you sir
love from india
Welcome to the channel!
thank u
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