Q.An ideal gas undergoes four different processes starting from the same initial state, drawn on a $P$-$V$ diagram (pressure $P$ vertical, volume $V$ horizontal). The four processes are, in some order, adiabatic, isothermal, isobaric and isochoric. Starting from the common initial point (upper-left region): curve 4 is the flat, topmost curve along which the pressure barely changes as the volume increases; curve 3 falls off below it as the volume increases; curve 2 is the lowest of the three expansion curves, falling most steeply as the volume increases; and curve 1 is a vertical line dropping straight down from the initial state (volume unchanged, pressure decreasing). Which numbered curve is the adiabatic process?

The fractional conversion of heat to work tells us what fraction of the heat supplied actually becomes useful work, rather than just raising the internal energy of the gas. In an isobaric (constant pressure) process, this fraction is determined entirely by the type of gas through its heat capacity ratio.

For any isobaric process, the first law gives us:

where $Q$ is heat supplied, $\Delta U$ is the change in internal energy, and $W$ is work done. The fractional conversion is:

Let me work through how this depends on the gas type.

1. Express work in an isobaric process

   At constant pressure, $W = P\Delta V = nR\Delta T$ for an ideal gas.

2. Express heat supplied

   For an isobaric process, $Q = nC_P\Delta T$, where $C_P$ is the molar heat capacity at constant pressure.

3. Calculate the fractional conversion

4. Relate to the heat capacity ratio

   We know that $C_P - C_V = R$ (Mayer's relation) and $\gamma = \frac{C_P}{C_V}$.

   From these: $C_P = \frac{\gamma R}{\gamma - 1}$

5. Substitute to find the fractional conversion

Now the key insight: this fraction increases as $\gamma$ decreases. The values of $\gamma$ for different gas types are:

Monatomic gases have the largest $\gamma$, which paradoxically gives them the highest fractional conversion because the formula is $1 - \frac{1}{\gamma}$, not proportional to $\gamma$ directly.

The concept first

For any quasi-static process, $W = \int P\,dV$ — literally the area under the curve on a $P$–$V$ plot. Go round a closed loop and the "areas under" the outward and return paths partly cancel; what survives is exactly the area enclosed:

The sign is set by the sense of travel: clockwise $\Rightarrow$ net work done BY the gas (positive), anticlockwise $\Rightarrow$ net work done ON the gas (negative). Also worth remembering: over a cycle $\Delta U = 0$, so $Q_{\text{net}} = W_{\text{net}}$ — a cycle simply shuttles heat into work (or vice-versa).

Step-by-step

1. Write the vertices.

2. Work on each leg.

• $C \to A$ (vertical, $V = V_1$ constant): isochoric, $dV = 0 \Rightarrow W_{CA} = 0$.
• $A \to B$ (horizontal, $P = P_1$ constant): isobaric expansion,

• $B \to C$ (straight slanted line, compression from $2V_1$ to $V_1$): the area under this line is that of a trapezium of parallel sides $P_1$ and $3P_1$ and width $V_1$:

and because the volume decreases, $W_{BC} = -2P_1V_1$.

3. Add them up.

4. Cross-check with the area rule. Base $=V_1$, height $=2P_1$:

and the traversal $C\to A\to B\to C$ is anticlockwise, so $W_{\text{net}} = -P_1V_1$. The two methods agree.

5. Read the options. The choices list magnitudes only; the magnitude of the net work is $P_1V_1$. (Physically: the gas absorbs net work $P_1V_1$ and rejects an equal amount of heat, since $\Delta U = 0$ over a cycle.)

ANSWER: B

Concept & Intuition

The gas supports the weight of chamber B plus the metal balls inside it. The pressure of the gas is simply the total weight supported divided by the area of the piston (chamber B’s base). Removing one ball reduces the total weight, so the pressure drops. The fractional change in pressure equals the fractional change in the supported weight, because the area stays constant. This is a direct application of $P = F/A$ for a gas in equilibrium with a movable piston.

1. Identify the forces in equilibrium Chamber B (with its n balls) rests on the gas. The gas exerts an upward force $P A$ (where $A$ is the cross‑sectional area of chamber B). This balances the total weight of chamber B plus the balls. Let $W_B$ be the weight of chamber B itself, and let $w$ be the weight of a single ball. Then

2. Pressure after removing one ball If one ball is taken away, the total weight becomes $W_B + (n-1)w$. The new pressure $P'$ satisfies

3. Find the difference $P - P'$ Subtract the two equations:

Hence

4. Express the original pressure $P$ in terms of $w$ and $A$ From the first equation,

But note that the weight of chamber B itself is also supported by the gas. However, the problem asks for $(P - P')/P$. We can compute this ratio without knowing $W_B$ separately if we realize that the change in pressure depends only on the weight of one ball, while the original pressure depends on the total weight. The ratio is

This does not simplify to a constant unless we know $W_B$. But wait — the problem statement says “the weight of chamber B is supported by the gas.” It does not say chamber B itself is weightless. So is the answer not simply $1/n$? Let’s re‑examine.

5. Assume chamber B is massless ($W_B = 0$) Then

So

For an adiabatic process no heat is exchanged ($Q = 0$), so by the first law of thermodynamics the work done by the gas equals the drop in its internal energy, $W = -\Delta U$. For an ideal gas this can be written entirely in terms of the end-state pressures and volumes:

Take state A as $(P_0, V_0)$. Using the adiabatic relation

to eliminate the unknown pressure at B, the work reduces to a multiple of $P_0 V_0$ over $(\gamma - 1)$ (equivalently over $(1-\gamma)$ once the overall sign is absorbed). Matching this reduced form against the given choices identifies option (B).