Animated Solution for Physics - Thermodynamics: The figure shows the $p-V$ plot of an ideal gas taken through a cycle $ABCDA$. The part $ABC$ is a semi-circle and $CDA$ is half of an ellipse. Then,

Imagine you are a thermodynamic detective, and you've just been handed a mysterious $p-V$ diagram. The graph shows an ideal gas undergoing a cyclic process $ABCDA$.

Our mission is to evaluate four specific claims about this cycle. Let's break down the physics step by step and uncover the truth behind each option.

Is it Isothermal?

Let's test the first claim. Is the path from $A$ to $B$ an isothermal process?

To answer this, we need to recall what an isothermal process looks like on a $p-V$ diagram. An isothermal process strictly follows Boyle's Law, which states that the product of pressure and volume is constant ($pV=\text{constant}$).

Geometrically, the graph of $pV=\text{constant}$ is a rectangular hyperbola.

Now, look closely at our path $A\to B$. It is explicitly given as a circular arc! Since a circular arc is not a rectangular hyperbola, the relationship between pressure and volume is not inversely proportional.

Therefore, the process $A\to B$ cannot be isothermal. Option (a) is incorrect.

Heat Flow

Moving on to the second claim, we need to determine if heat flows out of the gas during the path $B\to C\to D$.

To find the direction of heat flow, we must invoke the First Law of Thermodynamics:

First, let's look at the work done ($W$). As the gas moves from $B$ to $D$, its volume decreases from $V_B=3$ units to $V_D=1$ unit. Because the gas is being compressed ($\Delta V<0$), the work done by the gas is negative ($W<0$).

Next, we evaluate the change in internal energy ($\Delta U$). For an ideal gas, the internal energy is directly proportional to the absolute temperature, which in turn is proportional to the product of pressure and volume ($U\propto pV$).

Let's calculate this product at the endpoints: At point $B$: $p_B{V}_B=2\times 3=6$ At point $D$: $p_D{V}_D=2\times 1=2$

The product $pV$ has clearly decreased! This means the temperature has dropped, and consequently, the change in internal energy is negative ($\Delta U<0$).

Now, plug these back into the First Law. We are adding a negative work done to a negative change in internal energy. The result must be negative!

A negative $\Delta Q$ physically means that heat is flowing out of the gas. Thus, option (b) is absolutely correct.

Work Done in $A\to B\to C$

Let's evaluate the third claim: Is the work done during the path $A\to B\to C$ zero?

Remember, the work done by a gas is geometrically represented by the area under the $p-V$ curve.

From $A$ to $B$, the gas expands. The work done is positive, and it corresponds to the large area under the upper arc.

From $B$ to $C$, the gas compresses. The work done is negative, corresponding to the smaller area under the right-side curve.

When we add these together ($W_{ABC}=W_{AB}+W_{BC}$), the large positive area dominates the smaller negative area. The net area is clearly greater than zero.

Therefore, the net work done for this segment is positive, not zero. Option (c) is incorrect.

Net Work in the Cycle

Finally, let's look at the entire cycle $ABCDA$. What is the sign of the net work done?

There is a very elegant and simple rule for this. On any $p-V$ diagram, if a closed cycle is traced in a clockwise direction, the net work done by the gas is positive. This happens because the expansion phase occurs at higher pressures than the compression phase.

Conversely, a counter-clockwise cycle yields negative net work.

Our cycle $ABCDA$ is clearly moving in a clockwise direction. Therefore, the net work done by the gas in the complete cycle is positive. Option (d) is correct!