Q.An ideal gas is carried once around the cyclic process $A\to B\to C\to D\to A$ on a $P$-$V$ diagram whose four corners are: $A$ at (volume $V_o$, pressure $P_o$), $B$ at (volume $3V_o$, pressure $P_o$), $C$ at (volume $3V_o$, pressure $2P_o$) and $D$ at (volume $V_o$, pressure $2P_o$). The steps are: $A\to B$ an expansion at constant pressure $P_o$; $B\to C$ a pressure rise at constant volume $3V_o$; $C\to D$ a compression at constant pressure $2P_o$; $D\to A$ a pressure drop at constant volume $V_o$. Find the net work done by the gas in one complete cycle.

Both Assertion and Reason are false.

A cyclic process is, by definition, one in which the system undergoes a series of changes and returns exactly to its initial state (same pressure, volume, and temperature) at the end of the cycle.

Assertion (A): Internal energy U is a state function — it depends only on the state of the system, not on the path taken to reach it. Since a cyclic process ends in the same state it started in, ΔU_cycle = U_final − U_initial = 0 always, for any cyclic process. So the claim that internal energy 'always increases' in a cyclic process is false.

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Internal energy of a real gas = kinetic energy (depends only on T) + intermolecular potential energy (depends on molecular separation, i.e. on volume); compression reduces the separation, making the attractive PE more negative and thus lowering total internal energy at the same T.

For an ideal gas, internal energy depends only on temperature, since ideal-gas molecules are assumed to exert no forces on each other. But a real gas has weak attractive intermolecular forces between its molecules. This means the internal energy of a real gas is a function of both temperature and volume:

$U = U_{kinetic}(T) + U_{potential}(V)$

• The kinetic part depends only on $T$ (via the average molecular speed / equipartition of energy), so at the same temperature, this part is identical for both the compressed and the rarefied gas.
• The potential-energy part comes from the (attractive) intermolecular forces. When the gas is compressed, the average distance between molecules decreases, bringing them further within range of their mutual attraction. Since work must be done against attraction to pull molecules apart (and energy is released when they come closer under attraction), this potential energy becomes more negative as molecules get closer. …

For an ideal gas, internal energy depends only on temperature; since an isothermal process keeps temperature constant, ΔU = 0 during isothermal expansion.

The internal energy U of an ideal gas is a function of temperature alone (it does not depend on volume or pressure). This follows from the kinetic theory of gases, where the internal energy is entirely the kinetic energy of the randomly moving molecules, which depends only on T.

In an isothermal process, the temperature T of the gas is kept constant throughout. Since U = U(T) only, and T does not change, the internal energy of the gas also does not change:

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The first law of thermodynamics states that heat given to a system is used partly to increase its internal energy and partly to do external work: $\Delta Q = \Delta U + \Delta W$.

The first law of thermodynamics is essentially the law of conservation of energy extended to include heat as a form of energy transfer. If a quantity of heat $\Delta Q$ is supplied to a thermodynamic system, in general two things can happen: the internal energy of the system can change by $\Delta U$, and/or the system can do work $\Delta W$ on its surroundings (such as by expanding against external pressure). Since energy is conserved,

$\Delta Q = \Delta U + \Delta W$

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