Q.ii. A gas is allowed to expand in a well insulated container against a constant external pressure of 2.5 bar from an initial volume of 2.5 L to a final volume of 4.5 L. The change in internal energy, $\Delta U$ of the gas will be a. -500 J
b. + 500 J
c. -1013 J
d. + 1013 J

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).

Pressure and volume are free to change during an isothermal process -- only the temperature is held fixed. So options (b) and (c) describe isobaric and isochoric processes respectively, not isothermal.

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.
• Isothermal process: PV = constant (T fixed) - a downward-curving hyperbola (P decreases as V increases, not a straight/horizontal line).
• Adiabatic process: PV^gamma = constant - a steeper downward-curving hyperbola than the isothermal curve.

Since the given curve is a horizontal line (P essentially unchanged as V increases), it represents an isobaric (constant-pressure) process.

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.

(Note: the value gamma = 1.4 given in the question is a distractor relevant only to an ADIABATIC compression, where P V^gamma = constant would give a different factor of 2^1.4 approx 2.64 -- close to option (b)'s 2.8. But since the question explicitly says 'compressed isothermally', Boyle's law is the correct law to use, and the answer is exactly 2.)

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 $\Delta U = -w \neq 0$ in general; in isobaric/isochoric processes temperature usually changes, so $\Delta U \neq 0$.)