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Current Question (ID: 10989)

Question:

$\text{Thermodynamic processes are indicated in the following diagram. Match the following:}$ $\text{Column I: P. Process-I, Q. Process-II, R. Process-III, S. Process-IV}$ $\text{Column II: a. Adiabatic, b. Isobaric, c. Isochoric, d. Isothermal}$

Options:

1. $P \rightarrow a, Q \rightarrow c, R \rightarrow d, S \rightarrow b$
2. $P \rightarrow c, Q \rightarrow a, R \rightarrow d, S \rightarrow b$ (Correct)
3. $P \rightarrow c, Q \rightarrow d, R \rightarrow b, S \rightarrow a$
4. $P \rightarrow c, Q \rightarrow d, R \rightarrow b, S \rightarrow a$

Solution:

$\text{Hint: Recall the thermodynamic processes.}$ $\text{Step: Find the correct match.}$ $\text{In the case of the isochoric process, the graph is parallel to y-axis because the volume is constant. Therefore, the process I is isochoric.}$ $\text{In an adiabatic process, the slope of the graph depicting pressure versus volume (P-V graph) decreases sharply. This behavior indicates that, during an adiabatic expansion or compression, the temperature of the gas changes significantly due to the absence of heat exchange with the surroundings. Therefore, the process II is adiabatic.}$ $\text{Along the curve, the process is isothermal because the temperature is constant. Therefore, the process III is isothermal.}$ $\text{In the case of the isobaric process, the graph is parallel to x-axis because the pressure is constant. Therefore, the process IV is isobaric.}$ $\text{Therefore, the correct match is } P \rightarrow c, Q \rightarrow a, R \rightarrow d, S \rightarrow b.$ $\text{Hence, option (2) is the correct answer.}$

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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
}