An ideal gas is expanded from (p1, V1, T1) to (p2, V2, T2) under different conditions. The correct s

Q: Why is irreversible work represented by a rectangle on the P-V diagram?

Because the work done against a constant external pressure is $W = -P_{\text{ext}}\Delta V$. Graphically, this is the area of a rectangle with height $P_{\text{ext}}$ and width $\Delta V$.

Q: Why is the adiabatic curve steeper than the isothermal curve?

For an isothermal process, $P \propto 1/V$. For an adiabatic process, $P \propto 1/V^\gamma$. Since $\gamma > 1$ for all ideal gases, the pressure drops much faster for the same increase in volume in an adiabatic process.

Q: Does free expansion mean temperature always remains constant?

Only for an ideal gas! Because the internal energy of an ideal gas depends only on temperature. For real gases, cooling or heating can occur during free expansion due to intermolecular forces (Joule-Thomson effect).

Q: Why is $\Delta U$ negative in an adiabatic expansion?

In an adiabatic process, no heat enters or leaves the system ($q=0$). When the gas expands, it does positive work ($W>0$). According to the First Law ($\Delta U = q - W$), the gas must use its own internal energy to do this work, causing $\Delta U$ to be negative and the gas to cool down.

Q: Can a process be both isothermal and adiabatic at the same time?

Yes, the free expansion of an ideal gas is the classic example. Since it expands against a vacuum ($W=0$) and is insulated ($q=0$), its internal energy doesn't change ($\Delta U=0$). For an ideal gas, this means its temperature also doesn't change, making it isothermal.

Question

An ideal gas is expanded from $(p_1, V_1, T_1)$ to $(p_2, V_2, T_2)$ under different conditions. The correct statement(s) among the following is(are)

Options:

(A) The work done on the gas is maximum when it is compressed irreversibly from $(p_2, V_2)$ to $(p_1, V_1)$ against constant pressure $p_1$

(B) The work done on the gas is less when it is expanded reversibly from $V_1$ to $V_2$ under adiabatic conditions as compared to that when expanded reversibly from $V_1$ to $V_2$ under isothermal conditions.

(C) The change in internal energy of the gas (i) zero, if it is expanded reversibly with $T_1 = T_2$, and (ii) positive, if it is expanded reversibly under adiabatic conditions with $T_1 
eq T_2$

(D) If the expansion is carried out freely, it is simultaneously both isothermal as well as adiabatic.

Sigma Percentile · Accepted Answer

Let's dive into this beautiful thermodynamics problem. We have an ideal gas transitioning between states under different conditions. Our job is to act like detectives and verify four distinct claims about work and internal energy. Mastering these fundamental laws gives you absolute control over thermodynamics.

### Analyzing Option A: The Cost of Irreversibility

Let's test Option A. Imagine compressing the gas irreversibly from volume $V_2$ back to $V_1$ against a constant external pressure $P_1$. The work done **ON** the gas is simply the external pressure times the change in volume.

Mathematically, this is expressed as:
$$W_{	ext{irr}} = -P_{	ext{ext}} (V_1 - V_2) = P_1(V_2 - V_1)$$

On our P-V diagram, this is the area of the large rectangle bounded by $P_1$ and the volume limits. Now, if we compressed it reversibly, the work would be the area strictly under the curve. Clearly, the rectangular area for the irreversible process is much larger. So, the work done on the gas is indeed maximum when compressed irreversibly against $P_1$. Option A is absolutely correct.

### Analyzing Option B: The Steepness of the Adiabatic Path

Moving to Option B. Let's look at the expansion from $V_1$ to $V_2$. We have two paths: isothermal and adiabatic. Because the adiabatic process follows $PV^{\gamma} = 	ext{constant}$, its curve drops much faster. It is steeper than the isothermal curve, which follows $PV = 	ext{constant}$.

Work done **BY** the gas during expansion is the area under the P-V curve. Since the adiabatic curve lies below the isothermal curve, the area under it is smaller. Therefore, the work done in adiabatic expansion is less. Option B is spot on.

### Analyzing Option C: The Temperature Drop in Adiabatic Expansion

Let's evaluate Option C. Internal energy of an ideal gas depends exclusively on its temperature. In the first case, if it expands reversibly with $T_1 = T_2$, it is an isothermal process. No change in temperature means no change in internal energy. So, $\Delta U = 0$. The first part of statement C is correct.

But wait, look at the second part of Option C. For a reversible adiabatic expansion, heat exchange $q$ is zero. By the First Law of Thermodynamics:
$$\Delta U = q - W = -W$$

Since it is an expansion, work $W$ is positive, making $\Delta U$ negative. The gas does work at the expense of its own internal energy, so it cools down! But the statement claims $\Delta U$ is positive. That is a trap. Option C is incorrect.

### Analyzing Option D: The Paradox of Free Expansion

Finally, Option D. Free expansion means expanding against a vacuum, so external pressure is zero, and work done is zero ($W = 0$). If it is carried out freely in an insulated container, heat exchange is also zero ($q = 0$).

Therefore, by the First Law:
$$\Delta U = 0 - 0 = 0$$

For an ideal gas, internal energy depends only on temperature. So, $\Delta U = 0$ means temperature remains constant ($\Delta T = 0$). So, a free expansion is simultaneously adiabatic and isothermal. Option D is beautifully correct.

Animated Solution for Chemistry - Chemical Thermodynamics: An ideal gas is expanded from $(p_{1}, V_{1}, T_{1})$ to $(p_{2}, V_{2}, T_{2})$ under different conditions. The correct statement(s) among the following is(are)

Select Answer:

Visualized Solution

\text{Analyzing Thermodynamic Processes}

\text{Option A: Irreversible Compression}

\text{Option A: Reversible Compression}

\text{Option B: Expansion Curves}

\text{Option B: Work Comparison}

\text{Option C: Internal Energy (Isothermal)}

\text{Option C: Internal Energy (Adiabatic)}

\text{Option D: Free Expansion}

\text{Final Conclusion}

The Sigma Insight: First Law of Thermodynamics

Analyzing Option A

Analyzing Option B

Analyzing Option C

Analyzing Option D

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