You explain scientific concepts in such a brilliant and succinct way! I'm happy to have discovered your channel and I wish I had the pleasure of you teaching my classes :) Keep being awesome^^^
Nice.. since I knew that the enthalpy and internal energy of a gas are only a function of its temperature.. I was struggling to understand how the MechPE manual had an expression for reversible work for an isothermal process - my ?? was, where is the energy coming from to do work.. now I realize I was thinking of a closed system, and that was the error.. with an open system, work can be done or absorbed or by addition or subtrataction of heat through a heat sync .. certainly one of the more subtle/tricky concepts in thermodynamics (which is probably rarely utilized in practical applications) and your video cleared that up for me, thanks
Hi matey, thanks for the explanation what problem completely doll I cannot understand basic sing why if you’re applying heat temperature remain constant this I believe this is basic stink of iso thermic process but I cannot understand could you please explain
If I compress the piston, my work is converted into thermal energy inside the box. Then the heat sink takes that energy away. how can then the piston go back up, the energy should have been lost ?
As a bicycle pump inflates a tyre, it pressure rises from 30 kPa to 40 kPa at constant temperature of 30 °C. By assuming the air acts as an ideal gas, calculate the work done per mol of the air. A. -80.35 J B. 80.35 J C. -811.93 J D. 811.93 J (please show calculation) can use this formula W=nRT ln(p1/p2) Can you help me..everyone doesnt know how to solve this..i must present this question to my lect..please notice me
what if I want to do the isothermal curve on a P - v (specific volume) diagram? I don't know in witch table I should seek for the values to do the curve.
Can you help me? How much heat needs to be rejected during a true isothermal compression of 3,000 cubic feet of air, when compressed from 15 psi to 4,500 psi using a 3-stage compressor?
Thank you so much for explaining it so clearly, i was very lost until I found your video. Thank you very much!!
Finally found a satisfying explanation.
Thanks a lot, my friend.
I will recommend this channel to my friends too.
Subbed!
@@Elucyda you are very good but I didn't understood why is the internal energy (change) is zero.
You explain scientific concepts in such a brilliant and succinct way! I'm happy to have discovered your channel and I wish I had the pleasure of you teaching my classes :) Keep being awesome^^^
Nice.. since I knew that the enthalpy and internal energy of a gas are only a function of its temperature.. I was struggling to understand how the MechPE manual had an expression for reversible work for an isothermal process - my ?? was, where is the energy coming from to do work.. now I realize I was thinking of a closed system, and that was the error.. with an open system, work can be done or absorbed or by addition or subtrataction of heat through a heat sync .. certainly one of the more subtle/tricky concepts in thermodynamics (which is probably rarely utilized in practical applications) and your video cleared that up for me, thanks
I am from India
I am a student of class 11th
To which classes thermodynamics is most appropriate to teach in your country?
Finally found a good explanation. Thanks man!
I am indian but your explanation is so better
???
Hi! I'm a new fan.. Thank you for your lectures. I like the way how you explain things. Your voice is also very calm.
Very clear and informative thanks!
Thanku for teaching us in such a easy way..
You speak so clearly thanks. .well understood
Bro ur an amazing teacher ! Keep it up
Your explanations are amazing bro❤👍🏼 You know what you are saying 😂
Definitely deserves more subscribers
0:06
Thank you, I was having a confusion in that! You are amazing!
Subscribed
Your'e a great teacher!
nice .great explanation sir ,thank u .
Great video!
Nice explanation
You keep nice going 👏
Amazing!! helps me to understand easily😄
Thanks
Thank you. Sir
Nice video bro👍 well understood keep going 🙏
you're an angel thanks bro
Hi matey, thanks for the explanation what problem completely doll I cannot understand basic sing why if you’re applying heat temperature remain constant this I believe this is basic stink of iso thermic process but I cannot understand could you please explain
Now I understood the concept
If I compress the piston, my work is converted into thermal energy inside the box. Then the heat sink takes that energy away. how can then the piston go back up, the energy should have been lost ?
I know this every thing but i still can't understand how we can apply heat if temperature is constant
As a bicycle pump inflates a tyre, it pressure rises from 30 kPa to 40 kPa at constant temperature of 30 °C. By assuming the air acts as an ideal gas, calculate the work done per mol of the air.
A. -80.35 J
B. 80.35 J
C. -811.93 J
D. 811.93 J
(please show calculation)
can use this formula W=nRT ln(p1/p2)
Can you help me..everyone doesnt know how to solve this..i must present this question to my lect..please notice me
what if I want to do the isothermal curve on a P - v (specific volume) diagram? I don't know in witch table I should seek for the values to do the curve.
Baby brother of AK lectures.
Thank you for this.. 😊
Thank you so much!!!
Can you help me? How much heat needs to be rejected during a true isothermal compression of 3,000 cubic feet of air, when compressed from 15 psi to 4,500 psi using a 3-stage compressor?
Ooh thankyou so much 😍👍
Good men
Akoa aben😂. (Very intelligent)
Eu tentando estudar para termoquimica e voce apareceu kkk
Bro but in google Δu= Q-W so, when internal energy is zero, it becomes Q=W
Both sign conventions are correct as long as they are used consistently.
U r awesome
Here,∆U=0. Explain further.
thank uuuuu
❤️
Why not 𝑊=𝑛𝑅𝑇 ln (𝑉f/Vi)? Why have negative?
The negative sign comes from a sign convention for the work done. It is fine to use a different sign convention as long as you are consistent.