# U(internal energy)=nCvdT. why

1. Dec 20, 2011

### vkash

Internal of a gas is nCvdT. why it is nCvdT.do you know any simple proof or derivation for this.

Last edited: Dec 20, 2011
2. Dec 21, 2011

### Vagn

The heat capacity at constant volume is defined as the rate at which the heat changes with respect to temperature per mole. So for an infinitesimal change we can write.
$C_{v}=\frac{dQ}{dT}$

In an isovolumetric process no work is done so dU=δQ as per the 1st law of Thermodynamics
so we can write the equation as

$U=nC_{v}dT =n \frac{dQ}{dT}_{v}dT =n \frac{dU}{dT}_{v}dT$

Last edited: Dec 21, 2011
3. Dec 21, 2011

### technician

Cv is the molar heat capacity of a gas at constant volume and is defined as 'The heat energy required to warm 1 mole of a gas through one degree when its volume is kept constant'
Gases have 2 principal heat capacities. If the gas is kept at constant pressure then Cp is the molar heat capacity for gas at constant pressure.
When heat is supplied to a gas at constant volume no external work is done therefore all of the heat energy shows as a temperature change.
When heat is supplied to a gas at constant pressure some external work is done [P(V2 - V1)]
So for a temperature rise of 1 degree extra heat energy is required to provide the external work. This essentially means that Cp is greater than Cv and it can be shown that
Cp - Cv = R (the gas constant)

4. Dec 22, 2011

### vkash

great, but there is dent in this, that is if process is not isobaric(isovolumetric) then?????????
friend you seem to tell me that Cp-Cv=R. that is not what am i asking.

after all thanks to both guys,

I think i a just a beginner in thermodynamics. so proof of all these formula are out of my scope, I hope i will learn this formula in future.

5. Dec 22, 2011

### Morgoth

1st thermodynamic law:

δQ= dU + δW.
Supposing we have the general form of U=U(T,V)
then its differential:
dU=$\frac{\partial U(T,V)}{\partial T}$ dT + $\frac{\partial U(T,V)}{\partial V}$ dV

and the work is δW=pdV

we go to the 1st law and replace δW and dU by the quantities we have above. We get:

δQ=$\frac{\partial U(T,V)}{\partial T}$ dT + [$\frac{\partial U(T,V)}{\partial V}$ + p ] dV

in case of dV=0 (V:const) you get
$\frac{\partial U(T,V)}{\partial T}$ =δQ/dT $\equiv$ Cv

From that you totally see that:

U= n Cv dT (for n moles now)

of course that is for Cv constant, which of course is true for ideal gases.

Last edited: Dec 22, 2011
6. Dec 22, 2011

### technician

'Internal of a gas is nCvdT. why it is nCvdT.do you know any simple proof or derivation for this.'
The answer is :At constant volume no external work is done by (or on) the gas. Therefore the heat supplied = increase in internal energy.
To calculate the effect of heat supplied you nedd an 'SHC' equation
In general Heat energy = mass x SHC x temp change.
For a gas H = n x Cv x ΔT (n = number of moles rather than mass and C = molar heat capacity rather than specific heat capacity.... specific means 'per kg')

If you need to know the equation for when the pressure is kept constant you need a different C..... Cp. If you need something in between then you need something other than the principal Cv and Cp

7. Dec 22, 2011

### I like Serena

The general equation for dU in terms of temperature and volume is:
$$dU=n C_V dT + n \left[ T \left({\partial P \over \partial T}\right)_V - P \right] dV$$
(See wikipedia)
This result can be derived from the general formula dU=TdS-PdV, which is in terms of entropy and volume.
I'll leave the proof for that out (for now).

If you substitute the ideal gas law $P={nRT \over V}$, the requested result follows.

Last edited: Dec 22, 2011
8. Dec 22, 2011

### technician

In words:
Heat energy supplied = heat energy to raise temperature + heat energy converted to external work.(basically P x ΔV)
If there is no external work (constant vol) then
Heat energy = heat energy to raise temp