5 INTERNAL ENERGY CALCULATION | THERMODYNAMICS | IIT ADVANCED | JEE MAIN | CHEMISTRY CLASS 11

[@llight4619]

Lect 5 done....everytime someone likes this comment..i will rewatch it

[@llight4619]

Okay two likes!! I will rewatch it now,for revision!!

[@divyanshsingh5614]

lessgo@@llight4619

[@quasarmaverick]

@@llight4619 now 4 likes rewatch it😅

[@AmanRaj-bw3ug]

Re-watch it😂

[@pathfinderforsuccess]

Now 15th like

[@sarthakpriyadarshi3670]

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 🙏🙏🙏❤️❤️❤️❤️❤️

[@ArRt1719]

2:13 4:30 6:43 8:09 9:31 10:34 12:05 15:24

[@keshav409]

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.🙏

[@AHalo2019]

Leave that area

[@shauryaverma2_c_375]

Don't watch in 240p or u will go blind

[@DrGumskratch]

Who tf watches in 240p💀

[@Dev-i8r]

240 IS TOO MUCH WATCH AT 144

[@ahmedmalik-nc1ky]

💀​@@AHalo2019

[@schrodingersmiley6407]

No one has explained like you in whole youtube.

[@sohamunhale5606]

sir very nice lecture and explaination

[@MohitTyagi]

Keep watching

[@integration_IIT_ISM]

It was a mind blowing EXPLANATION...💥💥

[@deadpool20065]

Bro i love u 😽

[@ashishjoshi911]

@@deadpool20065 NO HOMO

[@imadiraj]

Love your lecture sir. ❤️❤️❤️❤️❤️❤️❤️❤️

[@kunals4509]

Sir what about Real Gas undergoing isobaric process?

[@nownow1025]

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.

[@archanagaikwad3063]

@@nownow1025 But why dp at last step

[@schrodingersmiley6407]

@@archanagaikwad3063 he has written by mistake it should not be there

[@deadpool20065]

@@nownow1025 absolutely right

[@azhar_writes_79]

Like that you are awesome 😎😎😎😎😎

[@vikram_singh005]

Why too much ads 😖😖

[@aadiparakh7225]

watching for midsems at bits used to watch for jee :)

[@PriyanshuKumar-dv5wl]

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?

[@comptech5240]

then ig u will use formula Q=U+W. Thus, as W is there, you will simply substract it from Q and get U.

[@comptech5240]

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.

[@apex_blade]

for an ideal gas U = f(t) since U = K.E and P.E = 0. hence work done doesn't matter

[@alphaiitd1]

Thanks sir

[@ramanujan4068]

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

[@study-y9w]

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.

[@KartikSharma-gb4uw]

Majja aa gya sir bhot acchi explanation thi ❤

[@lumapradhan8393]

Very nice sir

[@journeyofc6200]

Used to watch this for jee, now watching it for mid sem at NIT JSR😂

[@exteremethunder5211]

😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂

[@user-oh7be6vb5m]

sir how U is function of pressure please tell

[@jinnieslostalpaca6382]

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

[@infinitevoid5141]

Ideal gas eqn

[@edu_in_iitg]

Thank you sir

[@PavanVG-2506]

Sir when will you provide lectures in english If u provide it people from south get benefitted.....

[@jankiarora]

At 10:34 , for an ideal gas how is dU is zero?

[@phndkratv7498]

Gaseous state nhi padha kya

[@JaswanthSai_25]

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]

[@SSEpsilon]

Thank you sir

[@abhishekthakur3551]

Sir there are many ads between your lectures which disturbs the flow and especially there are ads of other coachings

[@MohitTyagi]

Dear student, if you are getting distracted by ads, you can join KZbin Premium membership or download the lecture videos to avoid distractions from ads.

[@sonalkumari3269]

Should we convert 20 degree celcius to 293K

[@Rohan-ob7es]

Not needed

[@mujhe_kya_mai_toh_tamatar_hu]

😊

[@amitsingh2782]

❤️❤️❤️

[@vikram_singh005]

🛐🛐🛐

[@frostgamerz2222]

#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

[@bhaskarraj378]

Thank you sir

[@namanarora2005]

Thank you sir