Q.The process in which the value of $\Delta U = 0$ is —

An isothermal process is, by its very definition (iso- = same, thermal = temperature), one in which the temperature of the system stays constant throughout.

In an isothermal process, the system is kept in thermal contact with a large reservoir (or changes occur slowly enough) so that temperature T remains unchanged even as pressure and volume vary (e.g. following PV = constant for an ideal gas, from PV = nRT with T fixed).

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A P-V graph tells you the process type from its shape: a straight horizontal line means constant pressure, i.e. an isobaric process.

The described figure shows a P-V graph with a nearly horizontal line: as volume (V, x-axis) increases, the pressure (P, y-axis) stays essentially constant.

Recall how each standard process appears on a P-V diagram:

• Isobaric process: P is constant while V changes - a horizontal straight line. This matches the figure.
• Isochoric process: V is constant while P changes - a vertical straight line. …

Because the process is explicitly isothermal (not adiabatic), Boyle's law applies directly: P1V1 = P2V2, giving a pressure increase factor of exactly 2, independent of gamma.

For an isothermal process on an ideal gas, temperature is constant, so Boyle's Law holds:

P1 V1 = P2 V2.

Given V2 = V1 / 2:

P2 = P1 V1 / V2 = P1 V1 / (V1/2) = 2 P1.

So the pressure increases by a factor of 2.

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Internal energy of an ideal gas depends only on temperature, so $\Delta U=0$ whenever $T$ is constant.

For an ideal gas, internal energy $U$ is a function of temperature alone. In an isothermal process, temperature stays constant throughout, so there is no change in internal energy, $\Delta U = 0$. (In an adiabatic process $q=0$ but …