In an isothermal process, how can the change in internal energy be 0?

Question

It was written in my textbook,

$$ \mathrm{d}U = \left(\frac{\partial U}{\partial T}\right)_V \mathrm{d}T +\left(\frac{\partial U}{\partial V}\right)_T \mathrm{d}V $$ If the process is isothermal, $\mathrm{d}T = 0$. So, the equation reduces to:

$$ \mathrm{d}U = \left(\frac{\partial U}{\partial V}\right)_T \mathrm{d}V $$

I was told, $\mathrm{d}U = 0$ in an isothermal process, does that mean $\mathrm{d}V = 0$, or $\left(\frac{\partial U}{\partial V}\right) = 0$? But how is it possible that $\mathrm{d}V=0$ in an isothermal process, or $\left(\frac{\partial U}{\partial V}\right) = 0$?

If someone could explain what am I doing wrong, that would be very helpful.

Answer 1

The internal energy of an ideal gas is simply $$ U = \alpha nRT,$$ where $$\alpha = \frac{\text{degrees of freedom}}{2}$$ So, in an isothermal process, $$ \Delta T = 0 \Longrightarrow \Delta U = \alpha nR\Delta T = 0,$$ and likewise any $$\left(\frac{\partial U}{\partial P}\right)_T = \left(\frac{\partial U}{\partial V}\right)_T = 0.$$

Answer 2

I was told, $\mathrm{d}U=0$ in isothermal process.

That is not generally true. It is, however, true for ideal gases, which is probably what you were discussing. No attractive or repulsive forces exist between ideal gas particles. Hence the only type of internal energy an ideal gas can have is kinetic energy, i.e., energy due to the motion of its particles. And since kinetic energy depends only on temperature, an ideal gas's internal energy likewise only depends on its temperature. As a consequence, $\mathrm{d}U = 0$ for ideal gases in isothermal processes, because the temperature doesn't change.

does that mean $\mathrm{d}V = 0$, or $\left(\frac{\partial U}{\partial V}\right) = 0$? But how is possible that $\mathrm{d}V=0$ in isothermal process, or $\left(\frac{\partial U}{\partial V}\right)=0$?

Because the internal energy of an ideal gas depends only on its temperature, $\left(\frac{\partial U}{\partial V}\right)_T=0$, i.e., the internal energy doesn't change with volume at constant $T$.

Answer 3

For completeness, since there is already a well-explained answer addressing how and why $\mathrm{d}U=0$ for an isothermal process is a hallmark of an ideal gas, here is a short derivation of a general expression for the energy.

Start from the total differential for the free energy: $$ \mathrm{d}U = \left(\frac{\partial U}{\partial V}\right)_T \mathrm{d}V +\left(\frac{\partial U}{\partial T}\right)_V \mathrm{d}T \tag{1}\label{eq:total-differential} $$

Define the heat capacity at constant volume as $$ C_V = \left(\frac{\partial U}{\partial T}\right)_V. \tag{2}\label{eq:heat-cap-const-v} $$

Evaluate the partial derivative with respect to $V$ in \eqref{eq:total-differential} from the 1st law of thermodynamics:

\begin{align} \mathrm{d}U &= -P\mathrm{d}V + T\mathrm{d}S \\ \rightarrow \left(\frac{\partial U}{\partial V}\right)_T &= -P + T\left(\frac{\partial S}{\partial V}\right)_T \tag{3}\label{eq:first-law} \end{align}

Make use of the following Maxwell relation: $$ \left(\frac{\partial S}{\partial V}\right)_T= \left(\frac{\partial P}{\partial T}\right)_V $$

So \eqref{eq:first-law} becomes \begin{align} \left(\frac{\partial U}{\partial V}\right)_T &= -P + T\left(\frac{\partial P}{\partial T}\right)_V \tag{4}\label{eq:with-Maxwell} \end{align}

Then inserting the results of \eqref{eq:heat-cap-const-v} and \eqref{eq:with-Maxwell} into \eqref{eq:total-differential} we obtain the general result (when only $pV$ work is done): $$ \mathrm{d}U = \left[ -P + T\left(\frac{\partial P}{\partial T}\right)_V \right] \mathrm{d}V + C_V \mathrm{d}T \tag{5} $$

If you plug in the equation of state for an ideal gas you then obtain $$\left(\frac{\partial U}{\partial V}\right)_T = 0$$ and the expected result $$\mathrm{d}U = C_V \mathrm{d}T.$$