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
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.
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.
@@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)😄
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????
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
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.
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.
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.
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.
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
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Why did you uses U=nCv*dt for an isobaric system. Cv is only used for iso-choric calculations right?
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
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.
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.
@@scottphan6172 but Cv and Cp has different values, right?
@@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)😄
Great work done!!
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????
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
Thank you so much 😢❤️
Thank you very ,uich sirrrrrr
Thank you so much!
Am a bit behind on the temperature conversion
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.
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.
why didn't you convert the 100 degree celcius to kelvins in the second question?
Can also we use Cv in question number 3 ?,sir
why when work is done by the system... i mean when gas expands why work is positive?
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
@@de-grafthazard9081 which is really F**king annoying.
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.
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.
hye, once i enter my degree life now I get what do you mean , thank you
how does 100 Celsius equal 100 kelvin....
It doesn't, but a CHANGE in Celsius is equal to a CHANGE in Kelvin. Very different.
@@mikevar9090 GG
How would you do the second question if it was a real gas?
Can this be used in chemistry?
what does it mean 0.08206 pleas could u tell me
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.
en.m.wikipedia.org/wiki/Gas_constant
3:41~part b
Why did you use 0.8206 instead of 8.3145 at 4:25?
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...
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
That's the coefficient of R when using atm for pressure and Liters for volume.
BTW it's 0.08206 atm Liter/(mole*K)
@@mikevar9090 what is atm?
and what is r?
How did you find R
R is the universal gas constant, 8.3145 joules per kelvin per mole
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