The way he teaches makes me smile .. and by seeing him smile it goes much wider .. Thanx sir for the efforts u making for the future generations 🙏🙏🙏❤️❤️❤️❤️❤️
Sir this vedio and vedios added after to this in thermodynamics aren't available in 240p pls 😭 pls make to them available in240p. I have net issues in my area.🙏
U (potential energy) depends on K.E. and P.E. which in turn depends on temperature(KE) and Volume as well as temperature (PE).....and Since volume is inversely proportional to pressure..... the dependence on Volume can also be considered dependence on Pressure
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10:06 U = f(ke) KE = 3 K T / 2 = 3 K (PV) / 2nR => ke = f(PV) = f (t) = f(p,v) = f(p,v,t) How did you say so. If T changes, then (p x v) is bound to change. Anybody, please clarify my philosphy if I'm wrong somewhere. How did you say that
The reasoning you're using involves the relationships between kinetic energy (KE), temperature (T), pressure (P), and volume (V), but there seems to be some confusion about how these quantities depend on each other in thermodynamic systems. Let’s break this down step by step and clarify where there might be confusion. ### 1. **Kinetic Energy and Temperature (KE = 3kT/2)** You’ve correctly mentioned that the average **kinetic energy** (KE) of a gas molecule in an ideal gas is proportional to the **temperature** (T): \[ KE = \frac{3}{2} k T \] where: - \( KE \) is the average kinetic energy per particle. - \( k \) is the Boltzmann constant. - \( T \) is the absolute temperature. This equation tells us that as the temperature \( T \) increases, the average kinetic energy of the gas molecules also increases. ### 2. **Relationship Between Temperature, Pressure, and Volume** You then mention another equation: \[ KE = \frac{3 k PV}{2nR} \] This expression comes from the **ideal gas law**: \[ PV = nRT \] where: - \( P \) is the pressure. - \( V \) is the volume. - \( n \) is the number of moles of gas. - \( R \) is the ideal gas constant. When we substitute \( T = \frac{PV}{nR} \) into the kinetic energy expression \( KE = \frac{3}{2} kT \), you get: \[ KE = \frac{3 k (PV)}{2nR} \] This simply shows that the kinetic energy can also be expressed in terms of pressure (P), volume (V), and temperature (T) via the ideal gas law. ### 3. **Dependence of KE on P, V, and T** You wrote: \[ KE = f(PV) = f(T) = f(P, V) = f(P, V, T) \] This expression is **not wrong** but can be misleading without further clarification. The kinetic energy is directly related to the **temperature** (T), and through the ideal gas law, it can also be written in terms of \( P \) and \( V \). However, in an ideal gas system, **if temperature (T) changes**, either \( P \) or \( V \) must change, or both, to satisfy the ideal gas law \( PV = nRT \). Therefore, \( KE \) is primarily a function of **T**, but since \( T \), \( P \), and \( V \) are related, any change in one will affect the others. ### 4. **Your Question about \( PV \) and T** You asked: > If \( T \) changes, then \( P \times V \) is bound to change. Yes, you are absolutely correct! According to the ideal gas law \( PV = nRT \), if the temperature \( T \) changes, either \( P \), \( V \), or both must change to maintain the relationship. Therefore, if \( T \) increases or decreases, the product \( P \times V \) must also adjust accordingly. ### Summary of Clarification: - The average kinetic energy \( KE \) of gas molecules is **directly proportional to temperature**. - The **ideal gas law** relates pressure (P), volume (V), and temperature (T) as \( PV = nRT \). - If **T changes**, then either \( P \) or \( V \), or both, will change to maintain the ideal gas law relationship. - Therefore, \( KE \) can be expressed as a function of \( T \) or indirectly as a function of \( P \times V \). Your philosophy about the relationship between these quantities is on point. If temperature changes, \( P \times V \) must change too-this is a direct consequence of the ideal gas law.
dU isnt zero for ideal gas. partial derivative of dU/dV is zero as the internal energy U = f(T) [Internal energy is function only in terms of T, so derivative with respect to V must be zero]
The way he teaches makes me smile .. and by seeing him smile it goes much wider ..
Thanx sir for the efforts u making for the future generations 🙏🙏🙏❤️❤️❤️❤️❤️
Lect 5 done....everytime someone likes this comment..i will rewatch it
Okay two likes!! I will rewatch it now,for revision!!
lessgo@@llight4619
@@llight4619 now 4 likes rewatch it😅
Re-watch it😂
Now 15th like
2:13 4:30 6:43 8:09 9:31 10:34 12:05 15:24
No one has explained like you in whole youtube.
It was a mind blowing EXPLANATION...💥💥
Bro i love u 😽
@@deadpool20065 NO HOMO
Sir this vedio and vedios added after to this in thermodynamics aren't available in 240p pls 😭 pls make to them available in240p.
I have net issues in my area.🙏
Leave that area
Don't watch in 240p or u will go blind
Who tf watches in 240p💀
240 IS TOO MUCH WATCH AT 144
💀@@AHalo2019
sir very nice lecture and explaination
Keep watching
Sir what about Real Gas undergoing isobaric process?
U=f(T,P) for real gas
dU={delU/delT}dT at const. P + {delU/delT}dT at const. T
phela term=0 since const. Pressure
dU={delU/delT}dT×dP at const. Temp.
@@nownow1025 But why dp at last step
@@archanagaikwad3063 he has written by mistake it should not be there
@@nownow1025 absolutely right
Love your lecture sir.
❤️❤️❤️❤️❤️❤️❤️❤️
In case 1
What if the system is doing some work(since it may not be isochoric) should we not consider work done by ideal gas?
then ig u will use formula Q=U+W. Thus, as W is there, you will simply substract it from Q and get U.
also ig as we are consider dT in the equation, all work done is to be considered by external energy supplied, no internal work done.
for an ideal gas U = f(t) since U = K.E and P.E = 0. hence work done doesn't matter
Like that you are awesome 😎😎😎😎😎
Thanks sir
watching for midsems at bits used to watch for jee :)
Thank you sir
Sir when will you provide lectures in english
If u provide it people from south get benefitted.....
Why too much ads 😖😖
Very nice sir
Majja aa gya sir bhot acchi explanation thi ❤
sir how U is function of pressure please tell
U (potential energy) depends on K.E. and P.E. which in turn depends on temperature(KE) and Volume as well as temperature (PE).....and Since volume is inversely proportional to pressure..... the dependence on Volume can also be considered dependence on Pressure
Ideal gas eqn
Should we convert 20 degree celcius to 293K
Not needed
Thank you sir
Sir there are many ads between your lectures which disturbs the flow and especially there are ads of other coachings
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😊
Used to watch this for jee, now watching it for mid sem at NIT JSR😂
That's a motivation for me❤
😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂
10:06
U = f(ke)
KE = 3 K T / 2 = 3 K (PV) / 2nR
=> ke = f(PV) = f (t) = f(p,v) = f(p,v,t)
How did you say so.
If T changes, then (p x v) is bound to change.
Anybody, please clarify my philosphy if I'm wrong somewhere.
How did you say that
The reasoning you're using involves the relationships between kinetic energy (KE), temperature (T), pressure (P), and volume (V), but there seems to be some confusion about how these quantities depend on each other in thermodynamic systems. Let’s break this down step by step and clarify where there might be confusion.
### 1. **Kinetic Energy and Temperature (KE = 3kT/2)**
You’ve correctly mentioned that the average **kinetic energy** (KE) of a gas molecule in an ideal gas is proportional to the **temperature** (T):
\[
KE = \frac{3}{2} k T
\]
where:
- \( KE \) is the average kinetic energy per particle.
- \( k \) is the Boltzmann constant.
- \( T \) is the absolute temperature.
This equation tells us that as the temperature \( T \) increases, the average kinetic energy of the gas molecules also increases.
### 2. **Relationship Between Temperature, Pressure, and Volume**
You then mention another equation:
\[
KE = \frac{3 k PV}{2nR}
\]
This expression comes from the **ideal gas law**:
\[
PV = nRT
\]
where:
- \( P \) is the pressure.
- \( V \) is the volume.
- \( n \) is the number of moles of gas.
- \( R \) is the ideal gas constant.
When we substitute \( T = \frac{PV}{nR} \) into the kinetic energy expression \( KE = \frac{3}{2} kT \), you get:
\[
KE = \frac{3 k (PV)}{2nR}
\]
This simply shows that the kinetic energy can also be expressed in terms of pressure (P), volume (V), and temperature (T) via the ideal gas law.
### 3. **Dependence of KE on P, V, and T**
You wrote:
\[
KE = f(PV) = f(T) = f(P, V) = f(P, V, T)
\]
This expression is **not wrong** but can be misleading without further clarification. The kinetic energy is directly related to the **temperature** (T), and through the ideal gas law, it can also be written in terms of \( P \) and \( V \).
However, in an ideal gas system, **if temperature (T) changes**, either \( P \) or \( V \) must change, or both, to satisfy the ideal gas law \( PV = nRT \). Therefore, \( KE \) is primarily a function of **T**, but since \( T \), \( P \), and \( V \) are related, any change in one will affect the others.
### 4. **Your Question about \( PV \) and T**
You asked:
> If \( T \) changes, then \( P \times V \) is bound to change.
Yes, you are absolutely correct! According to the ideal gas law \( PV = nRT \), if the temperature \( T \) changes, either \( P \), \( V \), or both must change to maintain the relationship. Therefore, if \( T \) increases or decreases, the product \( P \times V \) must also adjust accordingly.
### Summary of Clarification:
- The average kinetic energy \( KE \) of gas molecules is **directly proportional to temperature**.
- The **ideal gas law** relates pressure (P), volume (V), and temperature (T) as \( PV = nRT \).
- If **T changes**, then either \( P \) or \( V \), or both, will change to maintain the ideal gas law relationship.
- Therefore, \( KE \) can be expressed as a function of \( T \) or indirectly as a function of \( P \times V \).
Your philosophy about the relationship between these quantities is on point. If temperature changes, \( P \times V \) must change too-this is a direct consequence of the ideal gas law.
At 10:34 , for an ideal gas how is dU is zero?
Gaseous state nhi padha kya
dU isnt zero for ideal gas. partial derivative of dU/dV is zero as the internal energy U = f(T) [Internal energy is function only in terms of T, so derivative with respect to V must be zero]
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#askCompetishun sir apna kha PE=f(T,v) where v is displacement bw molecules but how if volume is 0 then v is also are they both same
Thank you sir
Thank you sir