Isobaric Process Thermodynamics - Work & Heat Energy, Molar Heat Capacity, & Internal Energy

Рет қаралды 165,628

6 жыл бұрын

This physics video tutorial provides a basic introduction into isobaric processes. It explains how to calculate the work done by a gas when it expands at constant pressure. It explains how to calculate the heat energy absorbed by a gas using the molar heat capacity at constant pressure and how to calculate the internal energy change of a gas by using the molar heat capacity at constant volume for monatomic gases and diatomic gases. This video includes a PV diagram for an isobaric process as well as the formulas and equations that go with it. This tutorial contains a few thermodynamics practice problems for you to work on.
Open Vs Closed Vs Isolated System:
• Open System, Closed Sy...
First Law of Thermodynamics:
• First Law of Thermodyn...
Isochoric Process:
• Isochoric Process Ther...
Isothermal Process:
• Isothermal process The...
Internal Energy of an Ideal Gas:
• Internal Energy of an ...
_________________________
Adiabatic Process:
• Adiabatic Process - Wo...
PV Diagrams:
• PV Diagrams, How To Ca...
Thermodynamics Review:
• Thermodynamics, PV Dia...
2nd Law of Thermodynamics:
• Second Law of Thermody...
Heat Engines:
• Heat Engines, Thermal ...
Converting Heat Into Electricity:
• Thermodynamics - Conve...
________________________
Carnot Cycle:
• Carnot Cycle & Heat En...
Otto Cycle:
• Otto Cycle of Internal...
Refrigerators and Heat Pumps:
• Refrigerators, Heat Pu...
Entropy:
• Entropy Change For Mel...
Heat Engines and Refrigerators Review:
• Carnot Heat Engines, E...
Physics PDF Worksheets:
www.video-tutor.net/physics-b...

Пікірлер: 42

@TheOrganicChemistryTutor 5 ай бұрын

Final Exams and Video Playlists: www.video-tutor.net/ Full-Length Math & Science Videos: www.patreon.com/mathsciencetutor/collections

@muhesipatrick5074 2 жыл бұрын

Great work done!!

@joanelumah8474 3 жыл бұрын

Thank you so much!

@tri_phobians7306 2 жыл бұрын

Thank you so much 😢❤️

@winproduction7585 2 жыл бұрын

Thank you very ,uich sirrrrrr

@gooddeedsleadto7499 Ай бұрын

In the case of Rankine cycle for Boiler or Refrigrstion heat is added or removed at constant pressure and there is no volume change. Specific volume change values when super heat is added or removed can be used to find the work? Thank u

@melikebuga6970 2 жыл бұрын

How would you do the second question if it was a real gas?

@amanuelgebreyohannes2896 4 жыл бұрын

Why did you uses U=nCv*dt for an isobaric system. Cv is only used for iso-choric calculations right?

@daltonbrewster8651 4 жыл бұрын

I think it's because in isochoric systems, there is 0 work done. Therefore, change in internal energy equals heat: U=Q-W --> U=Q=nCv*dt. It's just taking the same parameters from the given isobaric process and hypothetically simulating an isochoric process so that the work falls out and U is easily calculated

@scottphan6172 4 жыл бұрын

Change in internal energy is path independent. The equation is the same for any process, unlike Q and W which are path dependent and depend on what process the system is undergoing.

@SunSunSunn 3 жыл бұрын

I think it should've been Cp for an isobaric. Delta U = q + w -> dU = dq + dw. Since dH = dq for isobaric processes, we can take the derivative of dH = integrand Cp(T) dT to end up with Cp*dT.

@SSS20025 2 жыл бұрын

@@scottphan6172 but Cv and Cp has different values, right?

@soumyadiptabandyopadhyay9942 Жыл бұрын

@@SunSunSunn actually no, because the change in internal energy of an ideal gas is equal to the change in its kinetic energy. And K.E. = n(f/2R)T (Where f is the degrees of freedom and n is the total no. of moles) And (f/2R) is equal to Cv only. THE CHANGE IN INTERNAL ENERGY OF AN IDEAL GAS DEPENDS ONLY AND ONLY UPON TEMP. (it is independent of the type of process, it only depends on ∆T)😄

@georginaowusuakrasi7256 6 ай бұрын

Can this be used in chemistry?

@Sewe-nc7xl 2 ай бұрын

Can also we use Cv in question number 3 ?,sir

@chigo_e 4 жыл бұрын

Can someone help me with this question: 2 moles of an ideal gas kept inside a cylinder at an initial temperature of 27 degrees celsius is made to expand at a constant pressure of 2 x 10 to the sixth power until its volume is doubled. Calculate the work done by this isobaric process.

@gabor6259 4 жыл бұрын

W = n*R*deltaT Charles's law: V1/T1 = V2/T2. This gives us V1*T2 = V2*T1. The volume doubled, so V2 = 2*V1, if we substitute this back in the previous equation, we can cancel the V1 on both sides, and we get T1 = 2*T2, so the temperature (in kelvin) also doubled. 27 °C is 300 K, the double is 600 K, the change in temperature is deltaT = (600 - 300) K = 300 K. If you substitute this back in the first equation (along with the 2 moles), you should get approximately 5000 J.

@rosyy68532 Жыл бұрын

3:41~part b

@wawa99cute 5 жыл бұрын

why when work is done by the system... i mean when gas expands why work is positive?

@de-grafthazard9081 5 жыл бұрын

Syawl it's just a convention chosen by physicists. For Chemists, work done by a system is always negative but for physicists, it's positive

@Hyperion856 4 жыл бұрын

@@de-grafthazard9081 which is really F**king annoying.

@mikevar9090 3 жыл бұрын

Because in the Mechanical version of the first Law dW = +Pdv and is always positive for Work done BY the SYSTEM and Negative for Work Done ON the SYSTEM during an expansion. In the Chemistry Version of the First Law, the signs of the work are reversed for Work done BY and Work done ON the SYSTEM from the Mechanical Version.

@mikevar9090 3 жыл бұрын

In a constant pressure process: dWby the system = + PdV, or Wby = P (Vfinal -V Initial), since in an expansion Vf > Vi , you always get positive work of the gas pushing the piston out making the Volume larger. The gas is doing the work so it's Work by the system.

@wawa99cute 3 жыл бұрын

hye, once i enter my degree life now I get what do you mean , thank you

@mintetube391 Жыл бұрын

what does it mean 0.08206 pleas could u tell me

@janetanna8542 Жыл бұрын

As the units of volume and pressure are in Litre and atm respectively. 8.314 J is used when volume is in m^3 and pressure in Pascal.

@cbgaming08 9 ай бұрын

en.m.wikipedia.org/wiki/Gas_constant

@user-ri8mr8bx8f 4 ай бұрын

i love you

@SnaloShabalala 2 ай бұрын

How did you find R

@syahirahfarhanah707 2 ай бұрын

R is the universal gas constant, 8.3145 joules per kelvin per mole

@Pdq-_- 3 ай бұрын

@jorgeeduardocarrenozapata4243 3 жыл бұрын

how does 100 Celsius equal 100 kelvin....

@mikevar9090 3 жыл бұрын

It doesn't, but a CHANGE in Celsius is equal to a CHANGE in Kelvin. Very different.

@ZxNaba Ай бұрын

@@mikevar9090 GG

@MiiMaker 5 жыл бұрын

Why did you use 0.8206 instead of 8.3145 at 4:25?

@arkaroy213 5 жыл бұрын

Units provided litre...R is Universal Gas constant which varies according to unit 0.8206 L atm/Mol K.. 1.987 Cal/Mol K or 8.314 J/Mol K...

@215JoC 4 жыл бұрын

He did it to keep units constant. If the Volume is in Liters and your Pressure is in atm you will use R=0.08206 L/mol.K . If your Volume is in m^3 and your Pressure is in Pa ( which is N/m^2) then you use R=8.3145 J/mol.K

@mikevar9090 3 жыл бұрын

That's the coefficient of R when using atm for pressure and Liters for volume.

@mikevar9090 3 жыл бұрын

BTW it's 0.08206 atm Liter/(mole*K)

@dwaynecharmagnepedralba9458 2 жыл бұрын

@@mikevar9090 what is atm? and what is r?