I did this yesterday,i couldn't heed,and i wasted two hours of my life. Now i just got everything cleared in just a quarter in a foreign language. Yes you Khan ;)
You need to do a series on Quantum Thermodynamics. Classical Thermodynamics is like Classical Mechanics is to Quantum Mechanics. It's a mile-high view of something which is really working at the quantum level.
Sir namaste 🙏l am an Indian student. Before watching your videos my views on thermodynamics was so difficult subject but now it has been so easy for me. Thank you sir 😊
Henry Poincare named the conception of "entropy " as a " surprising abstract " Lev Landau wrote: " A question about the physical basis of the entropy monotonous increasing law remains open " One physicist said : " The entropy is only a shadow of energy“ # History: Clausius : dS= dQ /T. Boltzmann : S= k log W Planck : h*f = kT logW The formula of Entropy is : h*f = kT logW Israel Sadovnik Socratus
Entropy is everywhere even in our daily life. Just we don't realise it. For example, there are two libraries A and B. Both these libraries contains exactly same number of books. All the titles available in the library A is available in the library B as well. Imagine that the library A is well stacked and the library B is just a huge pile of books and all the books are just dumped. As per the first law of thermodynamics both these libraries are equal, (energy is conserved). But anyone who wants to take a book and do some reading will prefer library A only, which is well stacked and ordered ( Less entropy). Library A has less entropy and more order. While B has higher entropy and less order. We need to spend energy to bring it to the order. So basically entropy is a measure of disorder or randomness. Higher the disorder higher the entropy. We cannot measure/calculate entropy for all the process or in other words its very difficult to measure the entropy for some processes. There are equations for some standard engineering and chemical process to calculate entropy. If entropy associated with a process is very high,the process tend to have a low efficiency. Ref: Thermodynamics : An engineering approach. Y A Cengel (copied)
I have the biggest smile when It all clicked... !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! yes thank you thats why delta S gen is always positive. for a real process and its = 0 when Ideal !!!!!!!!!!!!!!!
Why do we even consider irreversibility to begin with, given entropy it is a state function & entropy for reversible as well as an irreversible cycle is the difference between the end points? Thanks
This is completely irrelevant but I was looking for information on entropy information theory and ended up here, and I just couldn't leave and I'm not sure if it was cause I was taking it what you were saying even though I havn't looked up much on previous related material, or if (most likely) it was your voice which seems to have me in some kind of calm/focus trance.
Does it mean change in entropy for an irreversible cycle is always equal to change in entropy due to internally irreversibility or friction in this case?
Cyclical integral dQ/T must always be greater than zero for an irreversible cycle because the total entropy is always increasing. Then, why the Clausius integral for an irreversible cycle is less than zero for an irreversible Cycle. Is it not kind of a contradiction? Could you please clarify?Thanks
Will the heat added from the friction not cause a temperature difference? Hence, the system will follow a different isotherm. How can the total entropy generation be calculated in a system with not-neglible resistance?
The heat generated by friction is NOT add to the system as energy, it is lost. So the internal energy lose its energy by the heat generated by friction.
@@ivankauf yes, Qa - Qr = 2Qf --> Qa = 2Qf + Qr, ie, the heat added is equal to the heat released plus all the heat lost to friction (which happens in two parts, one when going and another when going back).
2nd law of thermodynamics reaches when total equilibrium be reached. So energy of each space cube is equal to cubes in whole universe and dark energy is also spent out then universe might reach a temperature in which force of gravitational field is dominant. I think i am wrong.
Who are You man??? You teach physics, math, engineering materials, Madicine and history!!!! come on.... I think You are creative and unique and You must get an Oscar for all that videos and explaination and I think that i must kill my self soon cause i cant even understand my classes.... shittttt :( i just wont to say that You are a realley lead by example
@Chuichupachichi Sorry, dude. You are misinterpreting the laws of thermodynamics here. There is always an overall increase in entropy over the universe, but in subunits of space, there can be a decrease in entropy, such as the transfer of heat from a hot environment to the ultimately cold universe. When ice forms, there is a decrease of entropy within the ice. Therefore, much of what occurs on Earth has decrease in entropy when it cools, transferring the heat to the rest of the universe.
If you consider the Q in deltaS=Q/T to be the net heat ( Qsuba +Qsubf for expansion, -QsubR +Qsubf for contraction) then deltaS does equal zero, so what's the problem?
It is those who have interest come to the point.Good job bro.Anyway those not interested in this or in anything but only love themselves and their theories and say we have no right to call and correct them can be ignored or even given a piece of truth so this does not go bad like I hate the subject instead It was OK. Try on.I too am interested.Keep support up.
I don’t get something in the Carnot cycle. 1. Q1 and Q2 are transferred between two systems in the same temperature (the exterior and the gas), therefore there is no change in the entropy of the universe during these two processes, right? 2. In the remaining two processes of the Carnot cycle, there is no heat transfer, thus no change in the overall entropy again, right? So in the whole Carnot cycle, there seems to be no variation in the universe entropy, right? 3. How can there be no entropy variation if at the end, looking at the whole cycle, we have heat (Q1-Q2) going from a hotter source to a colder sink?
Gaudio Wind Remember that Q1 and Q2 are added and removed at different temperatures. Furthermore Q1/T1+Q2/T2=0 as the cycle goes back to the same point therefore net entropy change of the system is zero...Yes the entropy of universe is constant as there is no mechanism for entropy generation (all reversibilities are absent and no Q/T net is added)
Nikan RT Thanks a lot Nikan. It helped. Let me see if I got it straight. Let's suppose the hot source and cool sink are limited. Then after some time, that is, a limited number of cycles, T1 would decrease while T2 would increase and eventually would be the same, T1=T2, and the engine would have to stop, after all. I believe that at this moment, the overall entropy amount in the heat source and the sink would be higher then it was at the beginning of the process, right? Correct me if I am wrong. I believe that because I understand that when two bodies exchange heat to reach thermal equilibrium they have the overall entropy increased, right? And that would apparently be the same final state of the system source-sink, either it goes through the help of Carnot machine or just through a heat tunnel. Well, if the entropy of the universe remains the same, then the work extracted from the machine would have to decrease the entropy of the rest of the universe. Let's suppose the work was used to lift some weight. Does that make sense? The weight being higher would mean a decrease in the entropy of the system earth-weight?
Gaudio Wind When a cycle is completed, it goes back to the same point. Entropy is a state property so the total change in entropy for a completed cycle is zero. You are not supposed to connect the two reservoirs. If you did, the heat exchange between the two reservoirs would increase the entropy of the two reservoirs. The entropy of universe would increase.
Nikan RT Thanks again Nikan, but I think either I or you did not understand the overall situation. I think that only the gas comes to the same point at the end of each cycle. The hot source never gets back the heat it gave to the gas. And the cool sink never leaves back the heat it received from the gas. So a cycle is performed by the gas but not for the overall system constituted by the hot source, the cool sink and the gas. According to my understanding, during a complete cycle, heat is taken from the source, it is partially transformed into work and the rest is left to the sink. How can we think that the source and the sink come back to the same state?
Nikan RT Sorry, Nikan. I think I see what you mean now. And I Think I see my mistake. If the hot source and the cool sink are not ideal (or maybe not infinite) that is, if they could not keep their temperatures, the cycle could never occur, of course. There would be no those isothermal transformations, in the first place. I think we would have to execute a reverse cycle with another portion of gas at the same time, between the same source and sink, to keep their temperatures constant if they were limited. Does that make sense?
how u take temperature ( T ) because their is heat difference due to temperature difference , then why you take temperature as ( T ) ???? please reply ....
Friction has essentially three parts. The work or mechanical loss it creates, the heat produced that escapes the system, and the heat produced that stays in the system. You really should add a term for the work the friction force does to account for the mechanical energy loss instead of just rolling it into W in the equations. This makes me question how well you really understand thermodynamics.
I don't see how you're "clarifying the ... definition of entropy" because you've only explained that the *change* in entropy depends on the energy added. That doesn't explain what entropy actually is.
So if Q from friction remains, i get that it's not reversible, but since Q is energy, why is it bad since there is some extra energy. Is it because it's a parasite energy that's being taken away since it's friction, and friction is not good in an engine?
Pascu Ionut alright, let's consider a perfectly efficient system that is reversible. In that system, the work or energy you put in is able to be removed exactly equally. So if we had a frictionless piston and we pushed it with n force, we would get exactly n force back. However, in reality we don't have frictionless pistons. So in order for us to reverse the reaction, some heat is generated. Staying consistent with my previous analogy, if we were to push the piston with n force, in order to get it back, we would have to exert n+m force (where m is the extra force caused by friction that must be added in order to return the position to its initial state). So in terms of an engine, friction prevents us from getting back exactly the amount of energy we put in. I hope this helped somewhat.
KSI Foss what i ment was in the first equation, dUe = Qa - W + Qf , looks like the thermal energy from friction kind of works in our favor, and i know friction is not really wanted in engines. I just want to know if it.s true, as it seems from the plus sign, that the thermal energy from friction works in our favor...in theory at least, because it adds to more energy overall. is this correct or not? If i think about it, thermal energy from the friction between piston and cilinder is just the mechanical energy transformed into heat, but then again. if i look at the equation, the friction heat energy comes from nowhere, so it would be beneficial.
Pascu Ionut The best way to think about it in this case is that the energy in total is the energy required to do the reverse work on the system. Lets say then that we don't have any friction and a perfect system. that means that the forward reaction (Qa-W) plus the reverse reaction (-Qa+W) is equal to zero. Meaning that when we compress and release our hypothetical piston no energy is lost and it rebounds exactly to where it was. Now when we apply friction to the reverse reaction, instead of getting the sums of the two to equal zero, well get a slightly positive value for dUe, meaning that extra work is needed to reverse the reaction. This might help illustrate what i mean; Here's our equation with the friction added; dUe(forward)=Qa-W+F, and dUe(reverse)= -Qa+W+F. You see how the reverse reaction isn't the opposite as the forward when we add the friction in. Instead of the positive value of friction tuning negative in the reverse reaction, it stays positive! This means that over all it will take more work to reverse the reaction since the friction values don't cancel out. So in our engine, it requires more work to move the piston than what is theoretically possible. So the less friction we can muster, the more efficient the engine.
"Procyclic Heat Transformation Theory" remember that I have the (C) copywright on the theory....eventually its going to be known who actually gave you the idea. I know you proved the math, but nobody actually understanstands "Entropy" in a quantum leap.
+Mech E It's better to say that entropy is a measure of inaccessible energy (or not-done-work), rather than "entropy is unaccessible energy", since it is measured in different units, and it is decidedly not an energy.
you should also state that friction is another type of "inefficiency" which has NOTHING to do with the theoretical efficiency. To do a more correct version of this video, you should take rocks off, rather than pebbles (still frictionless). You should leave friction out of the discussion until the very end and state that real systems (engine), with friction cannot even get to the theoretical efficiency. Your use of friction in the videos is incorrect.
When a process takes place very fastly, we can APPROXIMATE it as adiabatic. For example, the compression and expansion processes in IC engines take place so fast that (approx. 40rps in petrol!) we can almost approximate it as adiabatic.
I did this yesterday,i couldn't heed,and i wasted two hours of my life.
Now i just got everything cleared in just a quarter in a foreign language.
Yes you Khan ;)
You need to do a series on Quantum Thermodynamics. Classical Thermodynamics is like Classical Mechanics is to Quantum Mechanics. It's a mile-high view of something which is really working at the quantum level.
This would have helped me a TON when I was studying thermodynamics in college. I'll be sharing this as much as I can.
Now you need not share
When Khan teaches he actual tries to bring some intuitive sense to the subject. I think this is the key to his genius.
Indeed
Sir namaste 🙏l am an Indian student.
Before watching your videos my views on thermodynamics was so difficult subject but now it has been so easy for me. Thank you sir 😊
these videos are really excellent, I only wish that they had numbers on them!!
The fact that I'm a 10th grader and can easily comprehend this just shows how great Sal is
Henry Poincare named the conception of "entropy "
as a " surprising abstract "
Lev Landau wrote:
" A question about the physical basis of the
entropy monotonous increasing law remains open "
One physicist said :
" The entropy is only a shadow of energy“
#
History:
Clausius : dS= dQ /T.
Boltzmann : S= k log W
Planck : h*f = kT logW
The formula of Entropy is : h*f = kT logW
Israel Sadovnik Socratus
Entropy is everywhere even in our daily life. Just we don't realise it.
For example, there are two libraries A and B. Both these libraries contains exactly same number of books. All the titles available in the library A is available in the library B as well. Imagine that the library A is well stacked and the library B is just a huge pile of books and all the books are just dumped.
As per the first law of thermodynamics both these libraries are equal, (energy is conserved). But anyone who wants to take a book and do some reading will prefer library A only, which is well stacked and ordered ( Less entropy).
Library A has less entropy and more order. While B has higher entropy and less order. We need to spend energy to bring it to the order.
So basically entropy is a measure of disorder or randomness. Higher the disorder higher the entropy. We cannot measure/calculate entropy for all the process or in other words its very difficult to measure the entropy for some processes. There are equations for some standard engineering and chemical process to calculate entropy.
If entropy associated with a process is very high,the process tend to have a low efficiency.
Ref: Thermodynamics : An engineering approach. Y A Cengel
(copied)
I really hate thermodynamics.
Lol issok
This is pure gold, Sir.
thanks a lot for clarifying the reversible process! My physics textbook sucks! It needs more examples.
I have the biggest smile when It all clicked... !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! yes thank you thats why delta S gen is always positive. for a real process and its = 0 when Ideal !!!!!!!!!!!!!!!
This clarified lot of thangss thankss
Thanks for your video. Very clear explanation.
Why do we even consider irreversibility to begin with, given entropy it is a state function & entropy for reversible as well as an irreversible cycle is the difference between the end points?
Thanks
Thanks
Thanks Alot. It helped alot.
Net heat added is -2qf and therefore delta s is negative
This is completely irrelevant but I was looking for information on entropy information theory and ended up here, and I just couldn't leave and I'm not sure if it was cause I was taking it what you were saying even though I havn't looked up much on previous related material, or if (most likely) it was your voice which seems to have me in some kind of calm/focus trance.
thank you. It helped
Does it mean change in entropy for an irreversible cycle is always equal to change in entropy due to internally irreversibility or friction in this case?
Cyclical integral dQ/T must always be greater than zero for an irreversible cycle because the total entropy is always increasing.
Then, why the Clausius integral for an irreversible cycle is less than zero for an irreversible Cycle.
Is it not kind of a contradiction? Could you please clarify?Thanks
Will the heat added from the friction not cause a temperature difference? Hence, the system will follow a different isotherm. How can the total entropy generation be calculated in a system with not-neglible resistance?
The heat generated by friction is NOT add to the system as energy, it is lost. So the internal energy lose its energy by the heat generated by friction.
At the end you have: Qa - Qr = 2Qf (considering Qf the heat lost by friction)
@@ivankauf yes, Qa - Qr = 2Qf --> Qa = 2Qf + Qr, ie, the heat added is equal to the heat released plus all the heat lost to friction (which happens in two parts, one when going and another when going back).
wondeful..perhaps the friction-generated heat is different when expanding and contracting?
i meant within the the water molecules that form the ice.
2nd law of thermodynamics reaches when total equilibrium be reached. So energy of each space cube is equal to cubes in whole universe and dark energy is also spent out then universe might reach a temperature in which force of gravitational field is dominant. I think i am wrong.
Who are You man??? You teach physics, math, engineering materials, Madicine and history!!!!
come on.... I think You are creative and unique
and You must get an Oscar for all that videos and explaination
and I think that i must kill my self soon cause i cant even understand my classes.... shittttt :(
i just wont to say that You are a realley lead by example
i wanna learn about constant volume
my fav uncle!
why is it that friction added only heat to the system and didn't add heat to that which the system was rubbing against to cause friction?
my brain started hurting by 1:41
@Chuichupachichi Sorry, dude. You are misinterpreting the laws of thermodynamics here. There is always an overall increase in entropy over the universe, but in subunits of space, there can be a decrease in entropy, such as the transfer of heat from a hot environment to the ultimately cold universe. When ice forms, there is a decrease of entropy within the ice. Therefore, much of what occurs on Earth has decrease in entropy when it cools, transferring the heat to the rest of the universe.
If you consider the Q in deltaS=Q/T to be the net heat ( Qsuba +Qsubf for expansion, -QsubR +Qsubf for contraction) then deltaS does equal zero, so what's the problem?
thermodynamics is awesome!
You and you only
It is those who have interest come to the point.Good job bro.Anyway those not interested in this or in anything but only love themselves and their theories and say we have no right to call and correct them can be ignored or even given a piece of truth so this does not go bad like I hate the subject instead It was OK. Try on.I too am interested.Keep support up.
I don’t get something in the Carnot cycle.
1. Q1 and Q2 are transferred between two systems in the same temperature (the exterior and the gas), therefore there is no change in the entropy of the universe during these two processes, right?
2. In the remaining two processes of the Carnot cycle, there is no heat transfer, thus no change in the overall entropy again, right? So in the whole Carnot cycle, there seems to be no variation in the universe entropy, right?
3. How can there be no entropy variation if at the end, looking at the whole cycle, we have heat (Q1-Q2) going from a hotter source to a colder sink?
Gaudio Wind Remember that Q1 and Q2 are added and removed at different temperatures. Furthermore Q1/T1+Q2/T2=0 as the cycle goes back to the same point therefore net entropy change of the system is zero...Yes the entropy of universe is constant as there is no mechanism for entropy generation (all reversibilities are absent and no Q/T net is added)
Nikan RT
Thanks
a lot Nikan. It helped. Let me see if I got it straight. Let's
suppose the hot source and cool sink are limited. Then after some
time, that is, a limited number of cycles, T1 would decrease while T2
would increase and eventually would be the same, T1=T2, and the
engine would have to stop, after all. I believe that at this moment,
the overall entropy amount in the heat source and the sink would be
higher then it was at the beginning of the process, right? Correct me
if I am wrong. I believe that because I understand that when two
bodies exchange heat to reach thermal equilibrium they have the
overall entropy increased, right? And that would apparently be the
same final state of the system source-sink, either it goes through
the help of Carnot machine or just through a heat tunnel. Well, if the
entropy of the universe remains the same, then the work extracted from
the machine would have to decrease the entropy of the rest of the
universe. Let's suppose the work was used to lift some weight. Does
that make sense? The weight being higher would mean a decrease in the
entropy of the system earth-weight?
Gaudio Wind When a cycle is completed, it goes back to the same point. Entropy is a state property so the total change in entropy for a completed cycle is zero. You are not supposed to connect the two reservoirs. If you did, the heat exchange between the two reservoirs would increase the entropy of the two reservoirs. The entropy of universe would increase.
Nikan RT
Thanks again Nikan, but I think either I or you did not understand the overall situation.
I think that only the gas comes to the same point at the end of each cycle. The hot source never gets back the heat it gave to the gas. And the cool sink never leaves back the heat it received from the gas. So a cycle is performed by the gas but not for the overall system constituted by the hot source, the cool sink and the gas. According to my understanding, during a complete cycle, heat is taken from the source, it is partially transformed into work and the rest is left to the sink.
How can we think that the source and the sink come back to the same state?
Nikan RT Sorry, Nikan. I think I see what you mean now. And I Think I see my mistake. If the hot source and the cool sink are not ideal (or maybe not infinite) that is, if they could not keep their temperatures, the cycle could never occur, of course. There would be no those isothermal transformations, in the first place. I think we would have to execute a reverse cycle with another portion of gas at the same time, between the same source and sink, to keep their temperatures constant if they were limited. Does that make sense?
This is why I choose to have dumb friends. It makes me think I'm intelligent.
That makes you look even dumber in reality.
thermodynamics is cool
wouldnt it be minus friction then? as it is basically heat generated by the system?
you're correct, heat released by the system is always negative
how u take temperature ( T ) because their is heat difference due to temperature difference , then why you take temperature as ( T ) ???? please reply ....
Friction has essentially three parts. The work or mechanical loss it creates, the heat produced that escapes the system, and the heat produced that stays in the system. You really should add a term for the work the friction force does to account for the mechanical energy loss instead of just rolling it into W in the equations. This makes me question how well you really understand thermodynamics.
Does anyone know what his input method is for all his videos, it doesnt seem like a mouse...
how the work w in compression and expension will in irreversible ,,,,,,???
is there any practical process where entropy is constant
It can not be constant, it always increase. we just assume it does not increase in order to compare between actual process and ideal process
@@التميمي-ض4ت i waited 8 years for this.
@@waqasaps
hhhhhhhhhhh
wow you are still alive man.
I hope you are a great engineer now (:
hahahah. yes i am 😊💯
@Winxin864 if you say so.
I don't see how you're "clarifying the ... definition of entropy" because you've only explained that the *change* in entropy depends on the energy added. That doesn't explain what entropy actually is.
So if Q from friction remains, i get that it's not reversible, but since Q is energy, why is it bad since there is some extra energy. Is it because it's a parasite energy that's being taken away since it's friction, and friction is not good in an engine?
Pascu Ionut alright, let's consider a perfectly efficient system that is reversible. In that system, the work or energy you put in is able to be removed exactly equally. So if we had a frictionless piston and we pushed it with n force, we would get exactly n force back. However, in reality we don't have frictionless pistons. So in order for us to reverse the reaction, some heat is generated. Staying consistent with my previous analogy, if we were to push the piston with n force, in order to get it back, we would have to exert n+m force (where m is the extra force caused by friction that must be added in order to return the position to its initial state). So in terms of an engine, friction prevents us from getting back exactly the amount of energy we put in. I hope this helped somewhat.
KSI Foss what i ment was in the first equation, dUe = Qa - W + Qf , looks like the thermal energy from friction kind of works in our favor, and i know friction is not really wanted in engines. I just want to know if it.s true, as it seems from the plus sign, that the thermal energy from friction works in our favor...in theory at least, because it adds to more energy overall. is this correct or not? If i think about it, thermal energy from the friction between piston and cilinder is just the mechanical energy transformed into heat, but then again. if i look at the equation, the friction heat energy comes from nowhere, so it would be beneficial.
Pascu Ionut The best way to think about it in this case is that the energy in total is the energy required to do the reverse work on the system. Lets say then that we don't have any friction and a perfect system. that means that the forward reaction (Qa-W) plus the reverse reaction (-Qa+W) is equal to zero. Meaning that when we compress and release our hypothetical piston no energy is lost and it rebounds exactly to where it was. Now when we apply friction to the reverse reaction, instead of getting the sums of the two to equal zero, well get a slightly positive value for dUe, meaning that extra work is needed to reverse the reaction. This might help illustrate what i mean; Here's our equation with the friction added; dUe(forward)=Qa-W+F, and dUe(reverse)= -Qa+W+F. You see how the reverse reaction isn't the opposite as the forward when we add the friction in. Instead of the positive value of friction tuning negative in the reverse reaction, it stays positive! This means that over all it will take more work to reverse the reaction since the friction values don't cancel out. So in our engine, it requires more work to move the piston than what is theoretically possible. So the less friction we can muster, the more efficient the engine.
"Procyclic Heat Transformation Theory" remember that I have the (C) copywright on the theory....eventually its going to be known who actually gave you the idea. I know you proved the math, but nobody actually understanstands "Entropy" in a quantum leap.
We know that more entropy means less useful work.........Can someone give me an example where the entropy increases and thus the work decreases?
+Mech E
It's better to say that entropy is a measure of inaccessible energy (or not-done-work), rather than "entropy is unaccessible energy", since it is measured in different units, and it is decidedly not an energy.
no cultural tolerance ftw?
Never said that.
entropy -or-Maxwell's Demon is the little carrot on a string out in front of the goat on the Little Rascals home made cart
There are no pain receptors in your brain therefore you can't feel pain.
why he didnt consider q2 as negative
because it's a loss heat
you should also state that friction is another type of "inefficiency" which has NOTHING to do with the theoretical efficiency. To do a more correct version of this video, you should take rocks off, rather than pebbles (still frictionless). You should leave friction out of the discussion until the very end and state that real systems (engine), with friction cannot even get to the theoretical efficiency. Your use of friction in the videos is incorrect.
and how does thermodynamics disprove Christianity?
@SteroidsR4success i don't think you understood my question.
Nothing can really be adiabatic?
I think so.
When a process takes place very fastly, we can APPROXIMATE it as adiabatic. For example, the compression and expansion processes in IC engines take place so fast that (approx. 40rps in petrol!) we can almost approximate it as adiabatic.
This proves my warm fart theory. I knew there was a good reason the warm ones smelled so bad. Q was written in brown. Awesome!
Please hindi me phadoooo fell nhi ata h
😂 haha
I'm the 115,001th viewer. :3
You are right! LOL
wiiiiiI!!!!
yo khan do u smoke weed lol?
Two Christians disliked this video.
I got a wife praise the Lord!!!