An ideal gas (0.5 mol), initially at 2 bar pressure, is compressed at a constant temperature of 600

Q: Why do we use $P_{\text{ext}}$ instead of the gas pressure for work?

In an irreversible process, the gas is not in equilibrium, so its internal pressure is not well-defined or uniform throughout the expansion/compression. The work exchanged with the surroundings is strictly determined by the external force (or pressure) acting on the piston.

Q: Why is the work done negative in the formula $W = -P\Delta V$?

This is the IUPAC sign convention. When a gas is compressed, $\Delta V$ is negative, so $-P\Delta V$ becomes positive, meaning work is done *on* the system, increasing its internal energy.

Q: How did we know the final volume of each step corresponds to the external pressure?

The problem states 'the compression is stopped only when the pressure of the gas becomes equal to the external pressure'. This means at the end of each step, the gas reaches mechanical equilibrium with the new external pressure.

Q: Is the geometric mean result ($P = \sqrt{P_1 P_3}$) always true for two-step isothermal compressions?

Yes! For any two-step isothermal compression or expansion between $P_1$ and $P_3$, the intermediate pressure that minimizes the total irreversible work (or maximizes work output in expansion) is always the geometric mean, $P = \sqrt{P_1 P_3}$.

Q: What if we used infinite steps instead of two?

If we used infinite infinitesimally small steps, the external pressure would always be equal to the internal pressure. The process would become reversible, and the work done would be the area exactly under the $P = nRT/V$ curve, which is the absolute minimum work possible: $W = -nRT \ln(P_1/P_2)$.

Question

An ideal gas ($0.5	ext{ mol}$), initially at $2	ext{ bar}$ pressure, is compressed at a constant temperature of $600	ext{ K}$ in two steps : first against a constant external pressure of $P	ext{ bar}$ ($2 < P < 8$), and then against constant external pressure of $8	ext{ bar}$. At each step, the compression is stopped only when the pressure of the gas becomes equal to the external pressure. The total work done on the gas in these steps is $W$. Considering all possible values of $P$ ($2 < P < 8$) and taking the gas constant as $R$ (in $	ext{J K}^{-1}	ext{ mol}^{-1}$), the minimum value of $|W|$ (in $	ext{J}$) is

Options:

(A) $207 R$

(B) $600 R$

(C) $630 R$

(D) $900 R$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you have a piston-cylinder arrangement filled with an ideal gas. Initially, the gas is relaxing at a pressure of $2	ext{ bar}$. We want to compress it to a final pressure of $8	ext{ bar}$, but we're going to do it in two distinct, sudden steps.

In the first step, we suddenly drop a weight on the piston, creating a constant external pressure $P$. The gas compresses until its internal pressure matches $P$. Then, we drop another weight, increasing the external pressure to $8	ext{ bar}$, and the gas compresses further until it reaches $8	ext{ bar}$. Throughout this entire process, we keep the cylinder in a large water bath at $600	ext{ K}$, ensuring the temperature remains perfectly constant. This is an isothermal process, but because we are changing the pressure in sudden jumps, it is highly irreversible.

### The Master Equation

For an irreversible compression against a constant external pressure, the work done on the gas is simply the external pressure multiplied by the change in volume. Since we have two steps, the total work $W$ is the sum of the work done in each step.

$$W = - P_{	ext{ext,1}} (V_2 - V_1) - P_{	ext{ext,2}} (V_3 - V_2)$$

Here, $P_{	ext{ext,1}}$ is our intermediate pressure $P$, and $P_{	ext{ext,2}}$ is our final pressure of $8	ext{ bar}$. But wait, we don't know the volumes! This is where the ideal gas law comes to the rescue. Since $PV = nRT$, we can express each volume as $V = \frac{nRT}{P_{	ext{internal}}}$.

Substituting these into our work equation gives us:

$$W = - P \left( \frac{nRT}{P} - \frac{nRT}{2} ight) - 8 \left( \frac{nRT}{8} - \frac{nRT}{P} ight)$$

### Simplifying the Math

Let's pull out the common factor $nRT$ to clean things up.

$$W = - nRT \left[ P \left( \frac{1}{P} - \frac{1}{2} ight) + 8 \left( \frac{1}{8} - \frac{1}{P} ight) ight]$$

Distributing the pressures inside the brackets, we get a beautiful simplification:

$$W = - nRT \left[ 1 - \frac{P}{2} + 1 - \frac{8}{P} ight]$$

$$W = - nRT \left[ 2 - \frac{P}{2} - \frac{8}{P} ight]$$

We are given $n = 0.5	ext{ moles}$ and $T = 600	ext{ K}$. So, $nRT = 0.5 	imes R 	imes 600 = 300R$. The question asks for the minimum value of the magnitude of work, $|W|$. Taking the absolute value flips the signs inside:

$$|W| = 300R \left[ \frac{P}{2} + \frac{8}{P} - 2 ight]$$

### The Optimization

To find the minimum work, we need to find the optimal intermediate pressure $P$. In calculus, we find minimums by taking the derivative and setting it to zero. Let's differentiate the term inside the bracket with respect to $P$:

$$\frac{d}{dP} \left( \frac{P}{2} + \frac{8}{P} - 2 ight) = \frac{1}{2} - \frac{8}{P^2}$$

Setting this to zero gives:

$$\frac{1}{2} = \frac{8}{P^2} \implies P^2 = 16 \implies P = 4 	ext{ bar}$$

So, the optimal intermediate pressure is exactly the geometric mean of the initial and final pressures ($\sqrt{2 	imes 8} = 4$). This is a classic result for two-step isothermal compressions!

### Final Calculation

Now, we just plug $P = 4$ back into our magnitude equation:

$$|W|_{	ext{min}} = 300R \left[ \frac{4}{2} + \frac{8}{4} - 2 ight]$$

$$|W|_{	ext{min}} = 300R [ 2 + 2 - 2 ] = 300R 	imes 2 = 600R$$

And there we have it! The minimum magnitude of work done on the gas is **$600R$**.

Select Answer:

Visualized Solution

Visualizing the Two-Step Compression

Work Done in Irreversible Compression

Applying the Ideal Gas Law

Simplifying the Work Expression

Substituting Known Values

Minimizing the Work Magnitude

Calculating the Minimum Work

The Way Forward

The Sigma Insight: First Law of Thermodynamics

Analyzing the Setup

The Master Equation

Simplifying the Math

The Optimization

Final Calculation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Two-Step Compression .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Applying the Ideal Gas Law .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Simplifying the Work Expression .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Minimizing the Work Magnitude .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating the Minimum Work .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

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