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@muhesipatrick50742 жыл бұрын
Great work done!!
@amanuelgebreyohannes28964 жыл бұрын
Why did you uses U=nCv*dt for an isobaric system. Cv is only used for iso-choric calculations right?
@daltonbrewster86514 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
@scottphan61724 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.
@SunSunSunn4 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.
@SSS200253 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)😄
@tri_phobians73062 жыл бұрын
Thank you so much 😢❤️
@kenny-oc4knАй бұрын
Am a bit behind on the temperature conversion
@chigo_e4 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.
@gabor62594 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.
@joanelumah84743 жыл бұрын
Thank you so much!
@winproduction75852 жыл бұрын
Thank you very ,uich sirrrrrr
@jorgeeduardocarrenozapata42433 жыл бұрын
how does 100 Celsius equal 100 kelvin....
@mikevar90903 жыл бұрын
It doesn't, but a CHANGE in Celsius is equal to a CHANGE in Kelvin. Very different.
@ZxNaba4 ай бұрын
@@mikevar9090 GG
@martinthomasvaz4608Ай бұрын
I have a doubt Sir, in question number 3 about calculating the change in internal energy of the system. The problem is about a constant pressure; volume expansion process, then why using Cv instead of Cp... any interpretation????
@gooddeedsleadto74994 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
@georginaowusuakrasi72569 ай бұрын
Can this be used in chemistry?
@wawa99cute5 жыл бұрын
why when work is done by the system... i mean when gas expands why work is positive?
@de-grafthazard90815 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
@Hyperion8565 жыл бұрын
@@de-grafthazard9081 which is really F**king annoying.
@mikevar90903 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.
@mikevar90903 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.
@wawa99cute3 жыл бұрын
hye, once i enter my degree life now I get what do you mean , thank you
@okocha_102 ай бұрын
why didn't you convert the 100 degree celcius to kelvins in the second question?
@melikebuga69703 жыл бұрын
How would you do the second question if it was a real gas?
@Sewe-nc7xl4 ай бұрын
Can also we use Cv in question number 3 ?,sir
@MiiMaker5 жыл бұрын
Why did you use 0.8206 instead of 8.3145 at 4:25?
@arkaroy2135 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...
@215JoC5 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
@mikevar90903 жыл бұрын
That's the coefficient of R when using atm for pressure and Liters for volume.
@mikevar90903 жыл бұрын
BTW it's 0.08206 atm Liter/(mole*K)
@dwaynecharmagnepedralba94583 жыл бұрын
@@mikevar9090 what is atm? and what is r?
@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.
@cbgaming0811 ай бұрын
en.m.wikipedia.org/wiki/Gas_constant
@rosyy68532 Жыл бұрын
3:41~part b
@SnaloShabalala5 ай бұрын
How did you find R
@syahirahfarhanah7075 ай бұрын
R is the universal gas constant, 8.3145 joules per kelvin per mole