Animated Solution for Physics - Thermodynamics: One mole of a monatomic ideal gas is taken through a cycle $ABCDA$ as shown in the $p-V$ diagram. Column II gives the characteristics involved in the cycle. Match them with each of the processes given in Column I.

A Journey Through the p-V Diagram

Thermodynamics is the study of energy in motion. When we look at a $p-V$ diagram, we are not just looking at lines and curves; we are looking at the heartbeat of an engine. Every cycle tells a story of heat entering, work being done, and energy being exhausted. In this problem, we are given a specific cycle $ABCDA$ for one mole of a monatomic ideal gas. Our mission is to decode this cycle, process by process, and match each leg of the journey with its physical characteristics.

Before we can analyze the thermodynamics, we must first understand the geometry of our cycle. The $p-V$ diagram provides us with the exact coordinates of each state. Let's extract them carefully.

State A is located at a volume of $3V$ and a pressure of $3p$. The product of pressure and volume here is $p_A{V}_A=9pV$.

State B is at a volume of $1V$ and a pressure of $3p$. The product is $p_B{V}_B=3pV$.

State C drops down to a pressure of $1p$ while maintaining a volume of $1V$. The product is $p_C{V}_C=1pV$.

State D expands to a volume of $9V$ at a constant pressure of $1p$. The product is $p_D{V}_D=9pV$.

Notice how the product $pV$ at state D is exactly the same as at state A. This is a crucial clue for the final leg of our cycle!

To determine whether internal energy increases or decreases, and whether heat is gained or lost, we need our master tools.

First, the internal energy $U$ of an ideal gas is given by $U=\frac{f}{2}nRT$. Using the ideal gas law $pV=nRT$, we can rewrite this as $U=\frac{f}{2}pV$. This tells us a beautiful secret: the internal energy is directly proportional to the product of pressure and volume. If $pV$ goes up, internal energy goes up. If $pV$ goes down, internal energy goes down.

Second, the First Law of Thermodynamics states that $Q=W+\Delta U$. The heat added to the system $Q$ is the sum of the work done by the gas $W$ and the change in its internal energy $\Delta U$.

Remember our sign conventions: Work $W$ is positive when the gas expands (volume increases) and negative when the gas is compressed (volume decreases).

Isobaric Compression

We start our journey from A to B. The pressure remains constant at $3p$, but the volume decreases from $3V$ to $1V$.

Since the volume is decreasing, the gas is being compressed. This means work is done on the gas, so $W<0$. This matches characteristic (t).

What about internal energy? The product $pV$ drops from $9pV$ at A to $3pV$ at B. Since $pV$ decreases, the internal energy must decrease, so $\Delta U<0$. This matches characteristic (p).

Now, let's look at heat. According to the First Law, $Q=W+\Delta U$. Since both $W$ and $\Delta U$ are negative, their sum $Q$ must also be negative. A negative $Q$ means heat is lost to the surroundings. This matches characteristic (r).

So, for process $A\to B$, the matches are p, r, and t.

Isochoric Pressure Drop

Next, we move from B to C. The volume is locked at $1V$, but the pressure drops from $3p$ to $1p$.

Because the volume doesn't change, the gas can't do any work. Therefore, $W=0$.

The product $pV$ drops from $3pV$ at B to $1pV$ at C. A decrease in $pV$ means a decrease in internal energy, so $\Delta U<0$. This matches characteristic (p).

Applying the First Law, $Q=0+\Delta U$. Since $\Delta U$ is negative, $Q$ is also negative. Heat is lost once again. This matches characteristic (r).

So, for process $B\to C$, the matches are p and r.

Isobaric Expansion

Now the gas fights back. From C to D, the pressure is constant at $1p$, but the volume expands massively from $1V$ to $9V$.

Since the volume is increasing, the gas is doing work on its surroundings. Therefore, $W>0$.

The product $pV$ increases from $1pV$ at C to $9pV$ at D. An increase in $pV$ means an increase in internal energy, so $\Delta U>0$. This matches characteristic (q).

With both positive work and positive change in internal energy, the First Law tells us that $Q=W+\Delta U$ must be positive. The gas is absorbing heat from its environment. This matches characteristic (s).

So, for process $C\to D$, the matches are q and s.

Isothermal Compression

Finally, we must return to our starting point to complete the cycle. The curve connects D to A.

Let's check the $pV$ products. At D, $pV=9pV$. At A, $pV=9pV$. The product is constant! For an ideal gas, a constant $pV$ means a constant temperature. This curve is an isotherm.

Because the temperature doesn't change, the internal energy remains perfectly constant. $\Delta U=0$.

The volume decreases from $9V$ to $3V$, meaning the gas is compressed. Work is done on the gas, so $W<0$. This matches characteristic (t).

Using the First Law, $Q=W+0$. Since $W$ is negative, $Q$ is also negative. Heat is lost to the surroundings. This matches characteristic (r).

So, for process $D\to A$, the matches are r and t.

By systematically applying the ideal gas law and the First Law of Thermodynamics, we have successfully decoded the entire cycle. We didn't need to memorize complex formulas for each specific process; we just followed the fundamental principles of energy conservation. This is the true power of thermodynamics!