thermodynamics
In isothermal process if the internal energy is not increasing then how does the pressure change?
I come to the question because in isobaric process the pressure remains unchanged because the change of volume and temperature . But in isothermal process the the temperature is not changed but pressure is changing.
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.
A P-V diagram is a good way to show what happens as the area under the graph is the work done and the gradient for an isothermal change is always larger (less negative) than that for an adiabatic change (the gradients are negative).
Compression from a volume $V_i$ to a volume $V_f$ results in a higher final pressure for an adiabatic change $P_{fa}$ than for an isothermal change$P_{fi}$ and hence more work is done during the adiabatic change.
Because the temperature has increased after the adiabatic change the internal energy of the system after the adiabatic change is greater than that after the isothermal change for which there is no change.
Best Answer
In isothermal process if the internal energy is not increasing then how does the pressure change?
It changes because in an isothermal process the product of pressure and volume is constant. For an ideal gas when heat is added to the system the system does an equal amount of work increasing in volume while decreasing in pressure. For an ideal gas where $PV=nRT$ this means temperature is a constant. Since the change in internal energy of an ideal gas depends only on temperature change, the change in internal energy is therefore zero. From the first law
$$\Delta U=Q-W$$
Since $\Delta U=0$, $Q=W$. For a reversible expansion the heat added to the system exactly equals the work done by the system, causing the volume to increase and pressure to decrease such that the product remains constant. For a compression the work done on the system exactly equals the heat leaving the system, causing the volume to decrease and pressure to increases so that the product is constant.
I come to the question because in isobaric process the pressure remains unchanged because the change of volume and temperature . But in isothermal process the the temperature is not changed but pressure is changing.
In an isobaric process the heat that is added to the system does work expanding the system but it also increases the internal energy of the system. It is because some of the heat goes into increasing the internal energy that the pressure can remain constant while volume and temperature increases.
Hope this helps.