Is Internal Energy of Ideal Gas Really Only Dependent on Temperature?

[s943035]

We known U=3/2nRT (monatomic ideal gas), just depends on temperature.

Most texts assert connecting U and Q with constant volume condition
and say"\Delta U = nCv\Delta T for any process because of internal energy only depends on temp".

I think that statement is very strange.
Deriving the first law to \Delta U = nCp\Delta T + nR \Delta T = n\Delta T (Cp + R). Well, this equation also just depends on temperature.

Why not both heat and work in other condition to determine internal energy? Maybe kinetic relation is derived from constant volume?

 

[Mapes]

Hi s943035, welcome to PF!

Where are you getting the equation \Delta U = nCp\Delta T + nR \Delta T = n\Delta T (Cp + R)? It does not seem correct. Maybe you mean \Delta U = nCp\Delta T - nR \Delta T = n\Delta T (Cp - R) for an ideal gas? This works out, since Cv = Cp-R.

 

[s943035]

Thanks for your greeting and correction!

So, we also can derive the internal energy relation from constant pressure and adapt for any process? Just texts favor start from constant volume?

Another thinking, why both energy relation (\Delta U = n\Delta T(Cv) = n\Delta T (Cp-R)) can adapt for any process?

If I assume we only remember PV^\gamma = C(C is constant, for adiabatic expansion) and forget about any internal energy description,
just from first law and ideal gas behavior in adiabatic expansion,
I got \Delta U = 0 - \int \frac{C}{V^\gamma}dv = -C\frac{V^{-\gamma +1}}{-\gamma +1} = C\frac{V^{-\frac{R}{Cv}}}{\frac{R}{Cv}}= C\frac{Cv V^{-\frac{R}{Cv}}}{R}
It seems like no any temperature relation, why?

 

[Mapes]

V is a function of temperature, so there is a temperature dependence. If you keep manipulating the equations, it should work out to give the same answer as before.

 

[s943035]

I don't know relation between V and T ...

 

[Mapes]

I don't know relation between V and T ...

V=nRT/P for an ideal gas.

 

[s943035]

V=nRT/P for an ideal gas.

but P is various in this case (adiabatic expansion),
using this relation substitute into, doesn't make wrong?

 

[Mapes]

I can see the equation getting pretty complicated, but in the end it's got to be equivalent to the simpler expression \Delta U=nc_V\Delta T.