Import Question JSON

Current Question (ID: 21056)

Question:

$\text{An ideal gas undergoes a cyclic transformation starting from the point A and coming back to the same point by tracing the path}$ $A \rightarrow B \rightarrow C \rightarrow D \rightarrow A$ $\text{as shown in the three cases below:}$

Options:

1. $\Delta U \text{ (Case-III)} > \Delta U \text{ (Case-II)} > \Delta U \text{ (Case-I)}$
2. $\Delta U \text{ (Case-I)} > \Delta U \text{ (Case-II)} > \Delta U \text{ (Case-III)}$
3. $\Delta U \text{ (Case-I)} > \Delta U \text{ (Case-III)} > \Delta U \text{ (Case-II)}$
4. $\Delta U \text{ (Case-I)} = \Delta U \text{ (Case-II)} = \Delta U \text{ (Case-III)}$

Solution:

$\text{Hint: Internal energy is a state function; its cyclic integral must be zero in a cyclic process.}$ $\text{For a cyclic process, the internal energy change } \Delta U \text{ is zero because internal energy (U) is a state function and depends only on the initial and final states.}$ $\text{Since the gas returns to the same state (A) in each case, we have}$ $\Delta U = U_{\text{final}} - U_{\text{initial}} = 0$ $\text{Since, } \Delta U = 0 \text{ for all three cases, we conclude } \Delta U \text{ case (I) = } \Delta U \text{ case (II) = } \Delta U \text{ case (III)}$ $\text{Hence, option 4 is the correct answer.}$

Import JSON File

Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}