**A very general discussion-Not specific to a system:**

The internal energy, $U$, of a system is a function of state, which means that its value only depends on the thermodynamic variables ($P, V, T)$ for example, at a given state (this means for a given set of values of these variables).

Let us make this more concrete: Imagine the system is in a thermodynamic state where the thermodynamic variables have the values ($P_i, V_i, T_i$) ($i$ stands for initial). At these values of the thermodynamic variables the internal energy has a value:

**Internal energy at the initial state $i$:** $U(P_i,T_i,V_i)$.

You can think of a gas at pressure, volume and temperature condition ($P_i, V_i, T_i$). Now imagine you change the thermodynamic variables to these ones ($P_f, V_f, T_f$) ($f$ stands for final). The internal energy now has a new value

**Internal energy at the initial state $f$:** $U(P_f,T_f,V_f)$.

In this process you have changed the internal energy of the system by an amount:

**Change in U:** $\Delta U= U(P_f,T_f,V_f)- U(P_i,T_i,V_i)$

I hope it is clear to observe that the system could have followed an infinitely large set of $(P,V,T)$-points, along an infinitely large number of different paths in order to go from state $i$ to state $f$. However, these are not, in any way, influencing by how much $U$ will change, you can take which ever path you please to go from state $i$ to state $f$. So the system has no memory of the intermediate states.

In mathematical terminology, this means that the differential change, $dU$, is a perfect differential and this is stated by the simple mathematical expression

$\oint_C dU=0$

It is very similar to the gravitational potential of the Earth, for example, which tells us that the amount of energy we need to spend to lift an object by 3m, does not depend whether we bring it straight vertically up or we follow some other path.

The quick answer is $\Delta U \neq 0$.

Let's look at some details.
In the special case where you are dealing with **ideal gas.** $$U = \frac{3}{2} nRT$$
Thus $$\Delta U = \frac{3}{2}nR\Delta T $$ Since the process is isothermal, $\Delta T$ is **zero**. Therefore $\Delta U = 0$. So it is not true that $q = 0$(that would be called adiabatic). Rather, $q = -w$.

The above analysis fails if the gas is **NOT** ideal. Since $U = \frac{3}{2}nRT$ is generally not true. But usually the ideal gas approximation works fine.

## Best Answer

Internal energy, pressure and volume are all state properties. A cycle returns all properties to their original values. So the change in each of these is zero. Enthalpy is defined as

$$H=U+PV$$

Therefore it too is a state property so its change is also zero.

Work and heat are energy transfers and are not properties. They depend on the processes carried out during the cycle. since $\Delta U=0$ then $W=Q$. Meaning the net work done in a cycle equals the net heat added.

Enthalpy is defined as above. It only equals the heat transferred for a constant pressure

process, as follows.$$\Delta H=\Delta U+\Delta (PV)$$ $$\Delta H=Q-W+P\Delta V+V\Delta P$$

For a constant pressure process, $W=P\Delta V$ and $V\Delta P=0$, and therefore

$$\Delta H=Q=mC_{p}\Delta T$$

And $\Delta H$ is positive if heat is added to the system resulting in a temperature increase.

But a process does not constitute a cycle.For a cycle the temperature must return to its original value. This require us to reverse the constant pressure process to transfer the same amount of heat out of the system making $\Delta H=0$ for the cycle.Hope this helps