Internal energy in irr. process with molar heat not constant

[Soren4]

I'm a bit confused about the following situation. In a irreversible thermodynamics process the molar heat of an ideal gas changes according to a function of the temperature, say $c_v=f(T)$ (which also leads to $c_p=R+f(T)$) and I'm asked to determine the heat exchanged during that process, knowing that (for istance) the process is isochoric.

In a isochoric process $\Delta U= Q$ but before to write this I need to find $\Delta U$, which is not $\Delta U= n c_v \Delta T$ because $c_v$ is not constant.

I would calculate it with the integral
$$\Delta U= \int dU= \int_{T_A}^{T_B} n f(T) dT$$

But this assumes that $dU$ can be written and integrated, which means that the temperature is defined in any intermediate step of the process, which is in contrast with the fact that the process is irreversible.

On the other hand I don't see another way to calculate the change in internal energy in such cases, so is this method correct? If it is not, then which is the correct method to get $\Delta U$ when $c_v=f(T)$ and the process is not reversible?

 

[Andrew Mason]

I'm a bit confused about the following situation. In a irreversible thermodynamics process the molar heat of an ideal gas changes according to a function of the temperature, say $c_v=f(T)$ (which also leads to $c_p=R+f(T)$)

Are you assuming an ideal gas?

and I'm asked to determine the heat exchanged during that process, knowing that (for istance) the process is isochoric.

In a isochoric process $\Delta U= Q$ but before to write this I need to find $\Delta U$, which is not $\Delta U= n c_v \Delta T$ because $c_v$ is not constant.

I would calculate it with the integral
$$\Delta U= \int dU= \int_{T_A}^{T_B} n f(T) dT$$

But this assumes that $dU$ can be written and integrated, which means that the temperature is defined in any intermediate step of the process, which is in contrast with the fact that the process is irreversible.

You only have to worry about work done on the surroundings. If the process is at constant volume, the only work done during the process is on the gas itself, not on the surroundings (eg. dynamic currents within the gas). When the gas returns to equilibrium, the work done on the gas simply contributes to internal energy (i.e. the kinetic energy of the currents is distributed and contributes to the average KE of all the molecules of the gas).

On the other hand I don't see another way to calculate the change in internal energy in such cases, so is this method correct? If it is not, then which is the correct method to get $\Delta U$ when $c_v=f(T)$ and the process is not reversible?

$\Delta U= \int dU= \int_{T_A}^{T_B} n f(T) dT$ is correct.

AM

 

[Chestermiller]

I agree with AM. U is a function only of temperature (for an ideal gas or an incompressible liquid or solid), and you are trying to find the change in U between the initial and final equilibrium states, irrespective of whether the process that took you from the initial state to the final state was irreversible. If I told you the initial temperature and final temperature in advance (and didn't reveal anything about how the system got from the initial to the final state), you would immediately calculate the change in U from the integral, with no hesitation. So, in short, it doesn't matter how the system got from the initial state to the final state, because U is a function only of state (and not path).