Physics 27 First Law of Thermodynamics (22 of 22) Work Done By A Gas

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In this video I will show you how to calculate the work done by a gas of isobaric, isothermic, and adiabatic process.

[@MichelvanBiezen]

Thank you all for comments. It is great to see that these videos are helping.

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[@MichelvanBiezen]

Thank you so much for your kind words. Glad you are enjoying the videos.

[@Skittix447]

Just got through your Thermodynamics playlist and I've got to say thank you. You've gotten me through my first semester of Physics and now my 2nd. Your videos are extremely helpful. THANKS

[@MichelvanBiezen]

Thanks for the feedback.

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I've watched all the 22 videos & I fully understand the whole subject now. Thank you, sir!!

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[@MichelvanBiezen]

mellow whlzki I am glad these videos are helping. Keep you your studies.

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Thank you. Really helpful ! I believe I've got 90% of understading. there other 10% is for practices.

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[@blueu127]

thank you soo much for this! i didnt understand the first law of thermodynamics, with the formulas changing depending on the state. i never knew i didnt have to memorize the formulas in order to understand. i just have to understand the concepts from your teaching. thank you professor!

[@MichelvanBiezen]

You're very welcome!

[@Reichvarg]

Thankyou for doing this series

[@AbigailSabandal]

Wow this playlist helped me a lot to understand these processes specially now that we are undergoing distance learning. I could not help but thank you, sir. You are an amazing educator.

[@MichelvanBiezen]

Thank you for writing. We are glad you find the videos helpful.

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[@MichelvanBiezen]

Timothy William Great! Thanks for letting me know.

[@Waleed9]

awesome sir!

[@fadlihafni3109]

thank you sir,you really helps me alot,which me luck on upcoming thermodynamic paper

[@MichelvanBiezen]

Good luck on your paper!

[@neworld2.073]

A steam turbine in a power station develops 1000 kW power. The amount of heat
supply into the boiler is 2800 kJ/kg, and the heat rejection from the condenser is
2100 kJ/kg. The power required to pump the steam to the boiler is 5 kW. Consider
the system is stationary and neglect the change of internal energies.
Applying the energy balance relation above, calculate the rate at which steam
circulates through the network of the power plant.

[@neworld2.073]

Could you help me with this if possible? Thanks

[@franksonbenedict544]

You deserve an award🎉

[@MichelvanBiezen]

Thank you. Your nice comment is the award! 🙂

[@peterv6i]

thank you!

[@pratyushdeshmukh7909]

Thanks a lot ......

[@MohammadkazimHalimy]

Thanks for your great explanation

[@MichelvanBiezen]

You are welcome!

[@nipunasnuwan8171]

there is a piston with steam cylinder,expands isothermaly and reversibly , in this case does the change of internal energy can zero or not? can u please help

[@MichelvanBiezen]

Isotherm, means the same temperature. If the temperature doesn't change then there cannot be a change in internal energy, since the internal energy only depends on temperature.

[@kevinmarquez117]

GOD BLESS YOU!

[@zahraasamer4411]

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[@paulmesler9412]

I am a little confused. In the first isobaric example, shouldn't you be using Cp, not Cv for the molar heat capacity in the delta U equation?

[@MichelvanBiezen]

+PR Mesler
To calculate the molar heat capacity for the delta U, you must always use Cv regardless of the process.

[@paulmesler9412]

+Michel van Biezen
Have you done a video explaining why Cv is always used when using delta U? I mean explaining in detail why this is the case because it's not intuitive to me.

[@MichelvanBiezen]

+PR Mesler
Watch the rest of the playlist and you'll get the information. I know this is a source of confusion because students confuse the Q in the first law of thermodynamics with the U. Note that U is the internal energy of the gas and it cannot depend on how you got there. Q is the heat gained or lost by the gas. The amount of heat does depend on the process during which the heat is exchanged. The best way to look at it is to take an isovolumetric process. No work is done because the volume of the gas doesn't change. Thus delta U equals Q which is n*Cv *delta T. In an isobaric process delta U is NOT equal to Q which is then n*Cp *delta T.

[@paulmesler9412]

+Michel van Biezen
So, are you saying that it doesn't matter if you have an isobaric, isovolumetric, isotherm,etc. process, delta U always uses Cv as the molar heat capacity?

[@MichelvanBiezen]

+PR Mesler That is correct.

[@miko74939]

Do I always need to convert liter to cubic meter with these processes?

[@MichelvanBiezen]

It is recommended

[@totallytechalicious]

Amazing series! :)

[@MichelvanBiezen]

Glad you think so!

[@lesterlozano4192]

Sir do you have video of polytropic process?

[@MichelvanBiezen]

No we don't

[@SergioPerez-tn8hs]

Thank you so much !!!!

[@MichelvanBiezen]

Glad to be of help. 🙂

[@sirisha890]

A vessel has two compartments of volume V1 and V2,containing an ideal gas at pressures P1 and P2 and temperatures T1 and T2.if the wall separating the compartments is removed,what will be resulting equilibrium temperature

[@MichelvanBiezen]

That is indeed an interesting problem. We are assuming here that both gases have the same specific heat? Then the final number of mols becomes nf= n1 + n2 and the final volume becomes Vf = V1 + V2. Thus Pf/Tf = nf * R / Vf, but that leaves you with a ratio. Next you can use the sum of the internal energies. U = n Cv T and thus Uf = n1 Cv T1 + n2 Cv T2 = U1 + U2 From here the final T can be calculated to be Tf = (n1 * T1 + n2 * T2) / (n1 + n2)

[@sirisha890]

Thankyou so much sirI

[@sirisha890]

Thankyou sir

[@shepherdzvavanjanja7253]

Great stuff

[@MichelvanBiezen]

Thank you for your comment. 🙂

[@asratnigusse3968]

Thank you very much

[@MichelvanBiezen]

You are welcome

[@kishen313]

respect ^^

[@nikhilstealth8486]

ISOVOLUMETRIC IS SAME AS ISOCHORIC RI8??

[@MichelvanBiezen]

+NIKHIL STEALTH
Yes, isovolumetric is the same as isochoric