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Current Question (ID: 8295)

Question:
$\text{Which of the following is correct for isothermal expansion of an ideal gas:}$
Options:
  • 1. $W_{\text{rev}} = W_{\text{irr}}$
  • 2. $W_{\text{rev}} + W_{\text{irr}} = 0$
  • 3. $W_{\text{rev}} > W_{\text{irr}}$
  • 4. $q_{\text{rev}} = q_{\text{irr}}$
Solution:
$\text{Hint: A reversible process (infinitely slow) does the maximum work.}$ $\text{Step 1: Understand the nature of reversible and irreversible processes}$ $\text{A reversible process is one that proceeds infinitesimally slowly through a continuous series of equilibrium states.}$ $\text{An irreversible process occurs at a finite rate and involves non-equilibrium states.}$ $\text{Step 2: Analyze work in expansion processes}$ $\text{For an expansion process (where the gas increases in volume), the gas does work on the surroundings.}$ $\text{By thermodynamic convention, work done by the system is negative.}$ $\text{For an isothermal expansion of an ideal gas:}$ $\text{• In a reversible process, the work done is: } W_{\text{rev}} = -nRT\ln\frac{V_2}{V_1}$ $\text{• In an irreversible process, the work done is typically: } W_{\text{irr}} = -P_{\text{ext}}(V_2 - V_1)$ $\text{Where } P_{\text{ext}} \text{ is the constant external pressure.}$ $\text{Step 3: Compare reversible and irreversible work}$ $\text{For the same initial and final states in an isothermal expansion:}$ $\text{• A reversible process extracts the maximum possible work from the system}$ $\text{• An irreversible process extracts less work due to entropy generation within the system}$ $\text{This means that for expansion processes (where work is negative):}$ $|W_{\text{rev}}| > |W_{\text{irr}}|$ $\text{Since both values are negative, and the reversible process has a larger magnitude:}$ $W_{\text{rev}} < W_{\text{irr}}$ $\text{However, in terms of the absolute amount of work extracted from the system (ignoring the sign convention):}$ $|W_{\text{rev}}| > |W_{\text{irr}}|$ $\text{Step 4: Consider the physical meaning}$ $\text{The reason a reversible process yields more work (in magnitude) is that:}$ $\text{• In an irreversible process, some energy is lost to entropy generation}$ $\text{• This entropy must be transferred to the surroundings as heat, reducing the amount of energy available to do work}$ $\text{• A reversible process has no entropy generation, maximizing the work output}$ $\text{For expansion, where the work is negative, a more negative value of } W_{\text{rev}} \text{ means more work is done by the system.}$ $\text{Therefore, for the isothermal expansion of an ideal gas:}$ $W_{\text{rev}} < W_{\text{irr}} \text{ (in terms of signed values)}$ $\text{This means more work is extracted in a reversible process.}$ $\text{In terms of the absolute magnitude of work (ignoring signs):}$ $|W_{\text{rev}}| > |W_{\text{irr}}|$ $\text{Looking at the options, } W_{\text{rev}} > W_{\text{irr}} \text{ is correct in terms of comparing the absolute amount of work done, since more work is extracted in a reversible process.}$ $\text{Option 3 is the correct answer.}$

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Expected JSON Format:

{
  "question": "The mass of carbon present in 0.5 mole of $\\mathrm{K}_4[\\mathrm{Fe(CN)}_6]$ is:",
  "options": [
    {
      "id": 1,
      "text": "1.8 g"
    },
    {
      "id": 2,
      "text": "18 g"
    },
    {
      "id": 3,
      "text": "3.6 g"
    },
    {
      "id": 4,
      "text": "36 g"
    }
  ],
  "solution": "\\begin{align}\n&\\text{Hint: Mole concept}\\\\\n&1 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\text{ moles of carbon atom}\\\\\n&0.5 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\times 0.5 \\text{ mol} = 3 \\text{ mol}\\\\\n&1 \\text{ mol of carbon} = 12 \\text{ g}\\\\\n&3 \\text{ mol carbon} = 12 \\times 3 = 36 \\text{ g}\\\\\n&\\text{Hence, 36 g mass of carbon present in 0.5 mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6].\n\\end{align}",
  "correct_answer": 4
}