I just want to take the time to tell you how much we appreciate you making these videos. + Just noticed that you have very few likes, then i remembered that i watched 25 videos in a row without liking so its nothing to be taken seriously! Will make sure to go back and like them though. Thanks again
So Just a heads up for people watching this IUPAC have really wacky definition of WORK and HEAT don't get it mixed up, professor here is not using IUPAC conventions.
Good point. There are 2 ways in which the first law of thermodynamics can be defined. The definitions used here is: The change in internal energy (delta U) = the heat added TO the gas - the work done BY the gas
You know there is probably not a better time to be a physics major. Reading the text is helpful, but watching you do examples is tremendously beneficial. I am a TA at my university. Would you mind if I included a link to your website on my TA page? I verbally tell students about your work all the time and tell them you're the only reason I am a successful physics major. Thanks for the continued help, professor!
I had a question similar to this in a recent exam. I failed it. It asked for the efficiency in one cycle (w/Q). While I can find w, Q is a mystery to me. I did the same thing for 1-2 and 3-1 but i just couldn't figure out 2-3. The professor has it done as Q= 3/2(P2v2-P1V1) + ((P1+P2)/2)*(V2-V1). I know this is a lot to ask but I would really appreciate it if you could explain. Thank you for your time Sir.
R = 8.3145 so it is rounded differently in different texts. You can measure the atmospheric pressure in 100 different places and you will never get 101,325 Pa. Taking into account that many insignificant figures is not necessary.
Sir, I have a question Did you explain in your chanile the “Steady State Flow” and some examples of Nozzle and Diffuser? If yes where can I found it? If no please explain it.
Work is the area under the process path on this graph. We have a trapezoid shape, if you extend the line of 2 to 3 down to zero to form this shape. The trapezoid reduces to a triangle for finding the net work of the full cycle. If you want the gross work of process 2 to 3, you would subtract the negative work of process 3 to 1.
When the volume increases, the system (gas) is doing work. (The gas is pushing against something). When the volume decreases, work is done on the system. (Something is pushing against the system (gas) and compressing it). The sign depends on how one defines the first law of thermodynamics.
The sign convention is arbitrary, and depends on the preferences of your audience. I recommend writing a comma clause with an interpretation of the sign in words. For instance, you'd say "W = -200J, negative sign indicates work is done on the system". You could lead to a misinterpreted answer if you write "W = -200J done on the system", so a separate clause or sentence will help clear that up. In thermodynamics problems in Physics, the sign convention is based on the system working like an engine. Heat is added to the engine, and work is done by the engine. Net work and net heat are both positive for a heat engine. Net work and net heat are both negative for a refrigeration cycle. Chemists tend to prefer the opposite sign convention for work, since that's the way IUPAC defines it. In mechanics, work is positive when done on an object, and negative when done by an object, because we want work to increase the KE of the system.
One last question ! I noticed that you don't always convert before solving for the change in, ex pressure or with liters, i know that the change in kelvins and celcius are the same, so slightly like that can we find the change in most or all of the variables (non si units ) before we convert ?
If the units given are not standard, and the calculation doesn't require a conversion to standard units, it seems more practical to keep the units as they are. But my advice always is, that if you are not sure if you should convert or not, convert to make sure you don't make a mistake.
Michel van Biezen thank you again! I was wondering since some equations that are entirely in non sI units and trying to write the sI units practically fills up all the space I have :D
You can use the Rankine scale as an alternative to the Kelvin scale, as long as you have consistent units in all other quantities. This can work as a shortcut if you are given Fahrenheit temperatures. Since both C and F have a possibility of being negative, you can't use them directly in the ideal gas law, or any equation that involves anything other than a temperature difference or a weighted average of temperatures. It would have the possibility of big problems when you divide by zero temperature, as well as the problem of a meaningless negative volume.
The 2-3 phase is not one of the 4 standard thermodynamic phases, but you can find the work done by finding the area under the curve, and you can find Q and delta U by calculating those values on the other 2 processes,
When I was calculating for Q and Change in U from state 3 to 1, I found Change in U (first) by using nCvChangeT, and then found Q, I looked back at how you calculated it and I find it really weird how I got the same answer, even though I used Cv since there wasn't constant volume? When I tried to use Cp (solving for Q first like in the video) , I got the same answerthough .., so I'm not sure how my answer even worked? Thanks!
Thank you very much! I'm studying for my final so all of these thermodynamic videos are extremely helpful (especially with conceptual problems that I usually don't understand).
You don't necessarily need constant volume for Cv to be an applicable term. All that matters for Cv instead of Cp, is that you are tracking the internal energy instead of the enthalpy. Any change in internal energy of an ideal gas, assuming Cv is constant, is: deltaU = n*Cv*(T2 - T1) And this is true whether the process is isobaric, isovolumetric, adiabatic, or anything else. The equation gets promoted to an integral, when Cv isn't constant with temperature. You can use Cv in a constant pressure process as long as you separately account for work. If you want heat to directly relate to change in temperature during a constant pressure process, then that is when you use Cp instead, and the corresponding concept of enthalpy (H).
+nikhil monarch You can graphically by calculating the area under the curve, just like you suggested. There just isn't a thermodynamic equation that describes that process (besides the area of a triangle) The change in the internal energy can be found when you know the initial and final temperature and when you know the type of gas and the amount of gas you have.
Good Day Sir, I have a similar problem to this but without numerical values. I have solved most of it and have got stuck at finding the efficiency. I just wanted to ask you if my chain of thought here is correct? eff = Work done / Heat in Q = heat (and 12,23,31 is the sides of the triangle) Since U = 0 , Q = W SO Q12+Q23+Q31 = W Heat is taken in in Q12 and Q23 and realased in Q31, and so Q12+Q23 = W - Q31 = Heat in eff = W/(W-Q31) Does that seem reasonable to you? :)
+Vital Soft The efficiency of a heat engine is: e = W/Qh which means that you only consider the heat taken in from the "hot" reservoir which is typically the power stroke of the engine.
The work done with area formulla is 2026J and when we solved the total work done by adding all the work done in individual processes , we came up with the ans 783J... why is that soo??plzz explain
We are not ignoring that area. The way it works is like this: from 1 to 2 there is no work done (no area under the curve). From 2 to 3 the work done is the entire area under the curve all the way down to the horizontal axis. Then the work done from 3 to 1 is negative work under the curve (volume is getting smaller so that work is done ON the gas NOT BY the gas). That area is therefore subtracted from the are under the curve from 2 to 3.
You cannot fine U from state 2 to state 3 but you can find delta U by subtracting the U at 2 from U at 3. You find W from 2 to 3 by subtracting W from 3 to 1 from the total W, etc.
@@MichelvanBiezen can we find the W from 2 to 3 by directly calculating the area under the slope? On Barron's AP I got this question wrong by doing so. I don't get it.
@@katerinapetrova7040 Indeed you can calculate W from 2 to 3, by the area under the slope, and extending P all the way down to zero. You can determine heat and change in internal energy, by inferring it from other information you have. In this problem, we know Q and deltaU for the process from 1 to 2, and for the process from 3 to 1, but not for 2 to 3. We know the changes in internal energy have to add up to zero (given that it is a complete cycle), which would allow us to solve for deltaU for the process from 2 to 3. Then simply apply Q = deltaU + W to get the heat associated with this process. Internal energy is a state function, so if you can determine temperature at any point along the way, you can determine the change in internal energy over any process. Heat and work are path functions, and require knowing the details of these processes.
On the T-S diagram, area corresponds to heat, rather than to work. You could find the heat associated with the process, and then use the state function of internal energy to determine deltaU for each process. With both these pieces of information, Q=deltaU + W allows us to solve for work.
It depends on how you define "W". Do you define it as "WORK DONE BY THE GAS"? or do you define it is "WORK DONE ON THE GAS". It is your choice. You are correct if you use the second definition, I use the first definition,
In an isothermal process, the internal energy of the gas doesn't change. Therefore all the heat needed for the gas to do work, has to be added to the gas.
Sir, how I will find efficiency of this cycle if there is no values given? Given only p1,p2,v1,v2 but no value of them. And could you plz upload a video about how we get different states graph from a given graph. for example, given a pv graph, now find out vt and pt graph.
+Md. Mominul Islam Note that in order to calculate the efficiency of the engine you must be told how much heat is added to the gas and how much is expelled in each cycle. Look at the videos in the playlist: PHYSICS 29 EFFICIENCY OF HEAT ENGINES
+Md. Mominul Islam The standard equation from which these graphs are derived is the ideal gas equation: PV = nRTYou define the vertical and horizontal axis and then you use the equation to find the relevant points on the graph.
In every cyclic process, the state goes back to the original state at the end of each cycle, and thus the state at the end is exactly the same as the state in the beginning, which means that the internal energy must be exactly the same. (but only at 1 point at the end of each cycle).
No. It is neither adiabatic nor isothermal. It is given that it is a straight line on the P-V diagram, and neither adiabatic nor isothermal are straight lines on the P-V diagram.
Michel van Biezen Damn, you actually replied. I was just simply commenting. Anyways your lectures are downright awesome sir. And replying even to my silly comments show how dedicated you are.
I just want to take the time to tell you how much we appreciate you making these videos.
+ Just noticed that you have very few likes, then i remembered that i watched 25 videos in a row without liking so its nothing to be taken seriously!
Will make sure to go back and like them though. Thanks again
So Just a heads up for people watching this IUPAC have really wacky definition of WORK and HEAT don't get it mixed up, professor here is not using IUPAC conventions.
Good point. There are 2 ways in which the first law of thermodynamics can be defined. The definitions used here is: The change in internal energy (delta U) = the heat added TO the gas - the work done BY the gas
You know there is probably not a better time to be a physics major. Reading the text is helpful, but watching you do examples is tremendously beneficial. I am a TA at my university. Would you mind if I included a link to your website on my TA page? I verbally tell students about your work all the time and tell them you're the only reason I am a successful physics major. Thanks for the continued help, professor!
Feel free to link. Glad to be part of the physics majors community. Thank you for spreading the word.
I'm from India and I'm IIT JAM PHYSICS exam aspirant...So thanks a lot......To covering my thermodynamic most important point.
Thank you very much.
my mans taught me more in 4 minutes than my teacher did in 3 weeks
Natural teacher
at 11:03 isnt Cp for monotonic gas 3/2 R instead of 5/2 R?
Cv = (3/2)R for a monatomic gas, but Cp = (5/2)R for a monatomic gas. Note that Cp = Cv + R
@@MichelvanBiezen thank you professor
woah so excellent
you explain what i never understood
Glad these videos are helping.
I had a question similar to this in a recent exam. I failed it. It asked for the efficiency in one cycle (w/Q). While I can find w, Q is a mystery to me. I did the same thing for 1-2 and 3-1 but i just couldn't figure out 2-3. The professor has it done as Q= 3/2(P2v2-P1V1) + ((P1+P2)/2)*(V2-V1). I know this is a lot to ask but I would really appreciate it if you could explain. Thank you for your time Sir.
The efficiency is defined as e = W/Qin = (Qin - Qout)/Qin
I always learned 1 atm as 101,325 Pa; R = 8.314. It would seem that there has been some adjustment of the constants.
R = 8.3145 so it is rounded differently in different texts. You can measure the atmospheric pressure in 100 different places and you will never get 101,325 Pa. Taking into account that many insignificant figures is not necessary.
Sir, I have a question
Did you explain in your chanile the “Steady State Flow” and some examples of Nozzle and Diffuser?
If yes where can I found it?
If no please explain it.
very helpful sir. thanks a lot !
I cannot understand the process of getting work done 2--3 ;
Work is the area under the process path on this graph. We have a trapezoid shape, if you extend the line of 2 to 3 down to zero to form this shape. The trapezoid reduces to a triangle for finding the net work of the full cycle. If you want the gross work of process 2 to 3, you would subtract the negative work of process 3 to 1.
Proffessor I'm confusid with the sign of workdone by the system and workdone on the system. In physics workdone on the system is Posistive? thank you
When the volume increases, the system (gas) is doing work. (The gas is pushing against something). When the volume decreases, work is done on the system. (Something is pushing against the system (gas) and compressing it). The sign depends on how one defines the first law of thermodynamics.
@@MichelvanBiezen I appreciate your politness
The sign convention is arbitrary, and depends on the preferences of your audience. I recommend writing a comma clause with an interpretation of the sign in words. For instance, you'd say "W = -200J, negative sign indicates work is done on the system". You could lead to a misinterpreted answer if you write "W = -200J done on the system", so a separate clause or sentence will help clear that up.
In thermodynamics problems in Physics, the sign convention is based on the system working like an engine. Heat is added to the engine, and work is done by the engine. Net work and net heat are both positive for a heat engine. Net work and net heat are both negative for a refrigeration cycle. Chemists tend to prefer the opposite sign convention for work, since that's the way IUPAC defines it.
In mechanics, work is positive when done on an object, and negative when done by an object, because we want work to increase the KE of the system.
One last question ! I noticed that you don't always convert before solving for the change in, ex pressure or with liters, i know that the change in kelvins and celcius are the same, so slightly like that can we find the change in most or all of the variables (non si units ) before we convert ?
If the units given are not standard, and the calculation doesn't require a conversion to standard units, it seems more practical to keep the units as they are. But my advice always is, that if you are not sure if you should convert or not, convert to make sure you don't make a mistake.
Michel van Biezen thank you again! I was wondering since some equations that are entirely in non sI units and trying to write the sI units practically fills up all the space I have :D
This playlist was made to show how every unit can be converted to the basis SI units, not that it is necessary to do so.
You can use the Rankine scale as an alternative to the Kelvin scale, as long as you have consistent units in all other quantities. This can work as a shortcut if you are given Fahrenheit temperatures.
Since both C and F have a possibility of being negative, you can't use them directly in the ideal gas law, or any equation that involves anything other than a temperature difference or a weighted average of temperatures. It would have the possibility of big problems when you divide by zero temperature, as well as the problem of a meaningless negative volume.
if phase 2-3 had a normal process how can you calculate the internal energy
The 2-3 phase is not one of the 4 standard thermodynamic phases, but you can find the work done by finding the area under the curve, and you can find Q and delta U by calculating those values on the other 2 processes,
When I was calculating for Q and Change in U from state 3 to 1, I found Change in U (first) by using nCvChangeT, and then found Q, I looked back at how you calculated it and I find it really weird how I got the same answer, even though I used Cv since there wasn't constant volume? When I tried to use Cp (solving for Q first like in the video) , I got the same answerthough .., so I'm not sure how my answer even worked? Thanks!
For the change in U you must always use Cv, regardless of the process.
Thank you very much! I'm studying for my final so all of these thermodynamic videos are extremely helpful (especially with conceptual problems that I usually don't understand).
@@MichelvanBiezen and if it is an isobaric process then will we not use Cp?
You don't necessarily need constant volume for Cv to be an applicable term. All that matters for Cv instead of Cp, is that you are tracking the internal energy instead of the enthalpy.
Any change in internal energy of an ideal gas, assuming Cv is constant, is:
deltaU = n*Cv*(T2 - T1)
And this is true whether the process is isobaric, isovolumetric, adiabatic, or anything else. The equation gets promoted to an integral, when Cv isn't constant with temperature.
You can use Cv in a constant pressure process as long as you separately account for work. If you want heat to directly relate to change in temperature during a constant pressure process, then that is when you use Cp instead, and the corresponding concept of enthalpy (H).
Sir can we find work done from 2 to 3 and internal energy of 2-3 ? If we can then how?
+nikhil monarch
You can graphically by calculating the area under the curve, just like you suggested. There just isn't a thermodynamic equation that describes that process (besides the area of a triangle)
The change in the internal energy can be found when you know the initial and final temperature and when you know the type of gas and the amount of gas you have.
Good Day Sir, I have a similar problem to this but without numerical values.
I have solved most of it and have got stuck at finding the efficiency.
I just wanted to ask you if my chain of thought here is correct?
eff = Work done / Heat in
Q = heat (and 12,23,31 is the sides of the triangle)
Since U = 0 , Q = W
SO Q12+Q23+Q31 = W
Heat is taken in in Q12 and Q23 and realased in Q31, and so Q12+Q23 = W - Q31 = Heat in
eff = W/(W-Q31)
Does that seem reasonable to you? :)
+Vital Soft
The efficiency of a heat engine is: e = W/Qh which means that you only consider the heat taken in from the "hot" reservoir which is typically the power stroke of the engine.
The work done with area formulla is 2026J and when we solved the total work done by adding all the work done in individual processes , we came up with the ans 783J... why is that soo??plzz explain
The answer from both he methods should be the same...
I also watched the (2of 4) video
why area under isobaric process not considered
We are not ignoring that area. The way it works is like this: from 1 to 2 there is no work done (no area under the curve). From 2 to 3 the work done is the entire area under the curve all the way down to the horizontal axis. Then the work done from 3 to 1 is negative work under the curve (volume is getting smaller so that work is done ON the gas NOT BY the gas). That area is therefore subtracted from the are under the curve from 2 to 3.
Michel van Biezen thank you sir
how i can find the W,Q,U for the state 2 to state 3
You cannot fine U from state 2 to state 3 but you can find delta U by subtracting the U at 2 from U at 3. You find W from 2 to 3 by subtracting W from 3 to 1 from the total W, etc.
@@MichelvanBiezen can we find the W from 2 to 3 by directly calculating the area under the slope? On Barron's AP I got this question wrong by doing so. I don't get it.
@@katerinapetrova7040 Indeed you can calculate W from 2 to 3, by the area under the slope, and extending P all the way down to zero. You can determine heat and change in internal energy, by inferring it from other information you have.
In this problem, we know Q and deltaU for the process from 1 to 2, and for the process from 3 to 1, but not for 2 to 3. We know the changes in internal energy have to add up to zero (given that it is a complete cycle), which would allow us to solve for deltaU for the process from 2 to 3. Then simply apply Q = deltaU + W to get the heat associated with this process.
Internal energy is a state function, so if you can determine temperature at any point along the way, you can determine the change in internal energy over any process. Heat and work are path functions, and require knowing the details of these processes.
could you please list to me where the internal energy U = zero ?
Not the internal energy, but the change in the internal energy is zero in a cyclic process.
How about, if triangle in TS diagram
On the T-S diagram, area corresponds to heat, rather than to work. You could find the heat associated with the process, and then use the state function of internal energy to determine deltaU for each process. With both these pieces of information, Q=deltaU + W allows us to solve for work.
PHY2AW I read that when v decreases w positive and vice-versa. So I think w 3-1 is positive?? Thank you
It depends on how you define "W". Do you define it as "WORK DONE BY THE GAS"? or do you define it is "WORK DONE ON THE GAS". It is your choice. You are correct if you use the second definition, I use the first definition,
Thank you for this.
what is the function of 2026J for both work and heat?
In an isothermal process, the internal energy of the gas doesn't change. Therefore all the heat needed for the gas to do work, has to be added to the gas.
during the state 1 to state 2 (Q=3039j & U=3039j) then cp=cv on this state????
There is no Cp from state 1 to state 2 since pressure is not constant. Thus Cv = delta U
Sir, how I will find efficiency of this cycle if there is no values given? Given only p1,p2,v1,v2 but no value of them. And could you plz upload a video about how we get different states graph from a given graph. for example, given a pv graph, now find out vt and pt graph.
+Md. Mominul Islam
Note that in order to calculate the efficiency of the engine you must be told how much heat is added to the gas and how much is expelled in each cycle. Look at the videos in the playlist: PHYSICS 29 EFFICIENCY OF HEAT ENGINES
Actually I watched this video to find efficiency of this unique system
Some years ago my teacher had explained it thoroughly and I knew how to find efficiency of this unique system. I was written on my notes.
MD sir if you know the solution for efficiency of this system. Please provide it to me.
Sir, could you please explain if a graph of PV is given, then how we find out the grph of VT & PV .
+Md. Mominul Islam The standard equation from which these graphs are derived is the ideal gas equation: PV = nRTYou define the vertical and horizontal axis and then you use the equation to find the relevant points on the graph.
@@MichelvanBiezen thank you sir
Hi professor plz why the internal energy not change ?
In every cyclic process, the state goes back to the original state at the end of each cycle, and thus the state at the end is exactly the same as the state in the beginning, which means that the internal energy must be exactly the same. (but only at 1 point at the end of each cycle).
isn't 2 to 3 adiabatic?
An adiabatic process has a steep curved slope, cutting through the isotherms.
No. It is neither adiabatic nor isothermal. It is given that it is a straight line on the P-V diagram, and neither adiabatic nor isothermal are straight lines on the P-V diagram.
Q=n*c(p)*(▲T) but you write Q=n*c(v)*(▲T) at 7:24
For an isovolumetric process (isochoric process), W = 0 and therefore delta U = Q - 0 and therefore Q = n Cv delta T
isovolumetric is called "isochoric"
Most text books are moving away from using isochoric and more typically use isovolumetric.
Michel van Biezen Damn, you actually replied. I was just simply commenting. Anyways your lectures are downright awesome sir. And replying even to my silly comments show how dedicated you are.
I appreciate viewers keeping me honest.
sir,
why T3 is 3 times T1?
Use PV = nRT If the volume triples and the pressure remains the same, the temperature must triple as well.
@@MichelvanBiezen ok thank you sir:)
sir,is it given that process from 2-3 is isothermic??
No, since it doesn't follow an isotherm (which has a curved appearance).From 2-3 is just a hypothetical process for the purpose of illustration.
Error: Monatomic gases have molar heat capacitie of 1.5R.
+k678kk
The video is correct.
Cv = 1.5 R Cp = 2.5 R
Michel van Biezen why do we use Cp here?
Sorry you're right.
JEE❤
Glad you like our videos.