[Hunt]

I remember from last semester that whenever we worked with ideal gases, we always took dU = nC_vdT for any process, and it never really did make much sense to me. Why should this hold true? I know for enthalpy dH = nC_pdT makes a lot more sense since dH = dq_p

Thanks in advance

[Donaldson Tan]

http://en.wikipedia.org/wiki/Kinetic_theory

You might find the Wikipedia article on Kinetic Theory of Gas useful.

The perfect gas is also assumed to adhere to the Kinetic Theory of Gas, and therefore internal energy of the perfect gas only takes account of translational and rotational kinetic energy.

It can be proven from the kinetic theory of gas that the translational and rotational kinetic energy is directly proportional to the temperature of the gas, ie. KE_{translational} = C₁T and KE_rotational = C₂T. Since internal energy (U) is a sum of translational and rotational kinetic energy, then U = KE_{translational} + KE_rotational = (C₁ + C₂).T . This implies that the internal energy of a perfect gas must be a function of its temperature.

Cv is defined as fixed-volume heat capacity.

Since we are considering the case for 1 mole of gas and there is no exchange of mass between the system and the surroundings, then this must be a close system.

Imagine we are heating 1 mole of perfect gas inside a rigid container, then dU = dQ + W

Since volume of the system is fixed, then dV = 0, so W = - p.dV = 0

This means dU = dQ + W = Q

heat capacity is defined as dQ/dT.

For an isochoric (fixed volume) process, dU = dQ, so the heat capacity = dQ/dT = dU/dT
=> Cv = dU/dT
=> dU = (dU/dT) dT = Cv.dT

[Hunt]

You might find the Wikipedia article on Kinetic Theory of Gas useful.

Thanks, we've already taken the derivation of the KMT. Certainly lots of properties or processes can be explained through it, but I fail to see how the expression dU=nC_vdT can be explained for any transformation.

The perfect gas is also assumed to adhere to the Kinetic Theory of Gas, and therefore internal energy of the perfect gas only takes account of translational and rotational kinetic energy.

It can be proven from the kinetic theory of gas that the translational and rotational kinetic energy is directly proportional to the temperature of the gas, ie. KEtranslational = C1T and KErotational = C2T. Since internal energy (U) is a sum of translational and rotational kinetic energy, then U = KEtranslational + KErotational = (C1 + C2).T . This implies that the internal energy of a perfect gas must be a function of its temperature.

I agree with you all the way.

Cv is defined as fixed-volume heat capacity.

Since we are considering the case for 1 mole of gas and there is no exchange of mass between the system and the surroundings, then this must be a close system.

Okay

Imagine we are heating 1 mole of perfect gas inside a rigid container, then dU = dQ + W

Since volume of the system is fixed, then dV = 0, so W = - p.dV = 0

This means dU = dQ + W = Q

heat capacity is defined as dQ/dT.

For an isochoric (fixed volume) process, dU = dQ, so the heat capacity = dQ/dT = dU/dT
=> Cv = dU/dT
=> dU = (dU/dT) dT = Cv.dT

You have considered a special case in which dV = 0 . The process is isochoric, and as you have just proven, dU=nC_vdT , but consider an isobaric process for instance, in which there's work done at const pressure by a specific expansion/compression, in this case dU = q - pdV , y would it be nC_vdT too? It makes no sense to me.

[Yggdrasil]

Most thermodynamic properties can be described as a function of two other thermodynamic properties.  Now, consider U as a function of temperature and volume:  U = U(T,V)

If we write the differential expression obtained from this function, we obtain:
dU = (dU/dT)_V dT + (dU/dV)_T dV
where (dU/dT)_V and (dU/dV)_T denote partial derivatives.

Now, note that (dU/dT)_V is the change in internal energy of the gas when temperature is varied at constant volume.  In other words, the partial derivative represents the molar constant volume heat capacity:
C_v,m = (dU/dT)_V.

Now consider (dU/dV)_T.  This is the change in internal energy of the gas when volume is varied at constant temperature.  Since internal energy reflects the potential energy of the gas, this term has to do with how the potential of the interactions between gas molecules changes as they move closer together or farther apart.  However, we assume that gas particles do not interact in an ideal gas, so (dU/dV)_T = 0.

Combining these two results with the differential we wrote above you get:

dU = C_v,mdT

[Donaldson Tan]

dU = (dU/dV)_TdV + (dU/dT)_VdT

http://en.wikipedia.org/wiki/Kinetic_theory
It can be proven from the kinetic theory of gas that the translational and rotational kinetic energy is directly proportional to the temperature of the gas, ie. KE_{translational} = C₁T and KE_rotational = C₂T. Since internal energy (U) is a sum of translational and rotational kinetic energy, then U = KE_{translational} + KE_rotational = (C₁ + C₂).T . This implies that the internal energy of a perfect gas must be a function of its temperature.

=> (dU/dV)_T = 0

=> dU = (dU/dV)_TdV + (dU/dT)_VdT = 0 + (dU/dT)_V = (dU/dT)_VdT = Cv.dT

[Hunt]

Thanks for your replies, Yggdrasil and geodome.

It makes a bit more sense now except that we've considered dU =f(T,V) , can we not express U as a function of temperature and pressure and end up with a different result, for instance? I mean, why volume? Perhaps temp was taken into consideration since U is theoretically directly proportional to the temp of the ideal gas, and volume was just a macroscopic property ?

[Donaldson Tan]

It makes a bit more sense now except that we've considered dU =f(T,V) , can we not express U as a function of temperature and pressure and end up with a different result, for instance? I mean, why volume? Perhaps temp was taken into consideration since U is theoretically directly proportional to the temp of the ideal gas, and volume was just a macroscopic property ?

Since internal energy of a perfect gas is only dependent on its temperature and not other parameters, such as pressure and volume, then dU/dP = dU/dV = 0. Only dU/dT is non-zero.

[Yggdrasil]

Thanks for your replies, Yggdrasil and geodome.

It makes a bit more sense now except that we've considered dU =f(T,V) , can we not express U as a function of temperature and pressure and end up with a different result, for instance? I mean, why volume? Perhaps temp was taken into consideration since U is theoretically directly proportional to the temp of the ideal gas, and volume was just a macroscopic property ?

Yes, you can say that U = U(T,P).  But then, for the differential you get:
dU = (dU/dT)_P dP + (dU/dP)_T dT
While (dU/dP)_T = 0 for an ideal gas, (dU/dT)_P has no physical meaning, so these are not convenient units to express internal energy.

However, despite their usefulness, T and V are not the most useful units for internal energy.  If you say that U = U(S,V) then you obtain one of the four fundamental equations of thermodynamics:
dU = TdS - PdV

[Hunt]

That's interesting ,? so if we assume U : --> U(S,V) , then :

dU = (dU/dS)_vdS for taking S as a variable.

and since dU = TdS - PdV = TdS

I end up with TdS = (dU/dS)_VdS

That is,? T = (dU/dS)_V.

Now, if we keep S const and take V as a variable, dU = -PdV

dU = (dU/dT)_SdT

P = - (dU/dV)_s ?

Does this make sense? Mathematically, it should , but if this is the case, we can prove other relations using other variables such as for U(P,T) , which should make physical sense ... I think.

[Donaldson Tan]

The whole idea of doing such derrivation is that we are interested how internal energy changes with entropy, volume, temperature, etc. If we can express this differentials in terms of measurable variables (eg. P,V,T) then we can measure the change easily. This set of equations is known as Maxwell Relations.

[Hunt]

Geodome, thanks a lot dude. I still have a few minor questions, but perhaps maybe later ... I'm satisfied for now.