One mole of an ideal gas at 900 K, undergoes two reversible processes, I followed by II, as shown be

Q: Why is Process I considered adiabatic?

Because it is a vertical line on the Entropy ($S$) axis, meaning the change in entropy $\Delta S$ is zero. A reversible isentropic process is, by definition, an adiabatic process where no heat is exchanged ($q=0$).

Q: Why does constant $U$ mean constant $T$ in Process II?

For an ideal gas, internal energy $U$ is solely a function of its kinetic energy, which is directly proportional to temperature ($U = n C_v T$). If $U$ doesn't change, $T$ cannot change, making it an isothermal process.

Q: Why did we use $W = \Delta U$ for Process I?

This comes directly from the First Law of Thermodynamics: $\Delta U = q + W$. Since Process I is adiabatic, the heat exchange $q$ is zero, leaving us with $W = \Delta U$.

Q: How did we know to use $C_{v,m} = \frac{5}{2}R$?

It was explicitly given in the problem statement! Always read the provided constants carefully; they often hint at the required mathematical path, in this case, finding the temperature from the internal energy.

Q: What is the sign convention for work used here?

We used the standard IUPAC chemistry convention where work done *on* the system is positive ($W = -\int P dV$). Since both processes are expansions (resulting in negative work), equating them directly works perfectly without sign errors.

Question

One mole of an ideal gas at $900	ext{ K}$, undergoes two reversible processes, I followed by II, as shown below.
If the work done by the gas in the two processes are same, the value of $\ln \frac{V_3}{V_2}$ is ___.

($U$: internal energy, $S$: entropy, $p$: pressure, $V$: volume, $R$: gas constant)
(Given: molar heat capacity at constant volume, $C_{V,m}$ of the gas is $\frac{5}{2}R$)

Sigma Percentile · Accepted Answer

### Decoding the U-S Diagram: A Journey Through Reversible Processes

Thermodynamics often tests our ability to translate abstract graphs into physical realities. In this problem, we are presented with a rather unusual indicator diagram: a plot of $\frac{U}{R}$ (Internal Energy scaled by the Gas Constant) against $S$ (Entropy).

Our mission is to decode two distinct thermodynamic processes and use a given constraint—that the work done in both processes is identical—to find a specific volume ratio.

### Phase 1: Decoding Process I (The Vertical Drop)

Let's direct our attention to Process I. Visually, it is a perfectly vertical line pointing downwards. What does a vertical line on an $S$-axis mean? It implies that the entropy of the system remains absolutely constant throughout the process, meaning $\Delta S = 0$.

In the realm of reversible thermodynamics, an isentropic (constant entropy) process is synonymous with an **adiabatic process**. This means there is zero heat exchange with the surroundings, so $q_I = 0$.

Armed with this knowledge, we invoke the First Law of Thermodynamics:

$$\Delta U = q + W$$

Since $q_I = 0$, the work done by the gas is simply the change in its internal energy:

$$W_I = \Delta U_I = U_2 - U_1$$

From the graph, we can read the initial and final states for Process I:
- Initial state: $\frac{U_1}{R} = 2250 \implies U_1 = 2250 R$
- Final state: $\frac{U_2}{R} = 450 \implies U_2 = 450 R$

Substituting these values, we find the work done in Process I:

$$W_I = 450 R - 2250 R = -1800 R$$

### Phase 2: Decoding Process II (The Horizontal Shift)

Now, let's analyze Process II. This process is represented by a horizontal line, which means the value on the y-axis, $\frac{U}{R}$, is constant. Consequently, the internal energy $U$ is constant.

For an ideal gas, internal energy is exclusively a function of temperature ($U = n C_v T$). If the internal energy doesn't change, the temperature cannot change. Therefore, Process II is a **reversible isothermal process**.

The standard formula for work done during a reversible isothermal expansion is:

$$W_{II} = -n R T_2 \ln\left(\frac{V_3}{V_2}ight)$$

We know $n = 1$ mole, but we are missing a critical piece of the puzzle: the temperature $T_2$ at the intermediate state.

### Phase 3: The Temperature Bridge

To find $T_2$, we must utilize the given molar heat capacity at constant volume, $C_{v,m} = \frac{5}{2}R$. The relationship between internal energy and temperature is:

$$U = n C_{v,m} T$$

Dividing both sides by $R$, we get:

$$\frac{U}{R} = \frac{n C_{v,m} T}{R}$$

At state 2, we know $\frac{U_2}{R} = 450$. Substituting our known values:

$$450 = \frac{1 	imes \left(\frac{5}{2}Right) 	imes T_2}{R}$$

$$450 = \frac{5}{2} T_2$$

Solving for $T_2$, we find:

$$T_2 = \frac{450 	imes 2}{5} = 180 	ext{ K}$$

Now we can complete our work equation for Process II:

$$W_{II} = -1 	imes R 	imes 180 	imes \ln\left(\frac{V_3}{V_2}ight) = -180 R \ln\left(\frac{V_3}{V_2}ight)$$

### Phase 4: The Grand Equivalence

The problem provides a master constraint: the work done in both processes is exactly the same.

$$W_I = W_{II}$$

We simply equate the two expressions we derived:

$$-1800 R = -180 R \ln\left(\frac{V_3}{V_2}ight)$$

The negative signs and the gas constant $R$ cancel out beautifully, leaving us with a straightforward algebraic equation:

$$\ln\left(\frac{V_3}{V_2}ight) = \frac{1800}{180}$$

$$\ln\left(\frac{V_3}{V_2}ight) = 10$$

And there we have it! By carefully translating the geometric features of the $U-S$ graph into thermodynamic principles, we arrived at a clean, integer solution.

Enter Numerical Value:

Visualized Solution

$U / R vs S Graph$

$Process I: Reversible Adiabatic$

$First Law for Process I$

$W_{I} Calculation$

$Process II: Reversible Isothermal$

$Work Done in Process II$

$Finding T_{2}$

$Substituting T_{2} into W_{I I}$

$Equating Works$

$Final Calculation$

$The Way Forward$

The Sigma Insight: First Law of Thermodynamics

Decoding the U-S Diagram

Phase 1

Phase 2

Phase 3

Phase 4

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } U/R vs S Graph

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Process I: Reversible Adiabatic

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } First Law for Process I

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } WI​ Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Process II: Reversible Isothermal

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Work Done in Process II

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Finding T2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting T2​ into WII​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equating Works

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward