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Current Question (ID: 8249)

Question:
$\text{Work done is maximum in -}$
Options:
  • 1. $\text{Isobaric work}$
  • 2. $\text{Isothermal work}$
  • 3. $\text{Isochoric work}$
  • 4. $\text{Adiabatic work}$
Solution:
\text{The maximum work is done when the external pressure } P_{ext} \text{ of the surroundings on the system is equal to } P\text{, the pressure of the system. If } V \text{ is the volume of the system, the work performed as the system moves from state 1 to 2 during an isobaric thermodynamic process, } W_{12}\text{, is the maximum work.} \text{Since for the same amount of expansion,} \text{Area}_{Isobaric} > \text{Area}_{Isothermal} > \text{Area}_{Adiabatic} > \text{Area}_{Isochoric} \text{Work done in thermodynamic processes is represented by the area under the P-V curve. From the P-V diagram shown:} - \text{Isobaric process: Constant pressure - represented by a horizontal line. Has the largest area under the curve for the same volume change.} - \text{Isothermal process: Constant temperature - represented by a hyperbolic curve } (PV = \text{constant})\text{.} - \text{Adiabatic process: No heat exchange - steeper curve than isothermal.} - \text{Isochoric process: Constant volume - vertical line, so area under curve = 0, hence no work done.} \text{For the same volume change, the isobaric process has the maximum area under the P-V curve, therefore maximum work is done.} \text{Therefore, option (1) is correct.}

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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
}