Import Question JSON

Current Question (ID: 8332)

Question:

$\text{Match the following process with entropy change:}$

Options:

1. $\begin{array}{|c|c|c|c|} \hline & A & B & C \\ \hline 1. & 2 & 3 & 1 \\ \hline 2. & 1 & 2 & 3 \\ \hline 3. & 1 & 3 & 2 \\ \hline 4. & 3 & 2 & 1 \\ \hline \end{array}$
2. $\begin{array}{|c|c|c|c|} \hline & A & B & C \\ \hline 1. & 1 & 2 & 3 \\ \hline 2. & 1 & 2 & 3 \\ \hline 3. & 1 & 3 & 2 \\ \hline 4. & 3 & 2 & 1 \\ \hline \end{array}$
3. $\begin{array}{|c|c|c|c|} \hline & A & B & C \\ \hline 1. & 1 & 3 & 2 \\ \hline 2. & 1 & 2 & 3 \\ \hline 3. & 1 & 3 & 2 \\ \hline 4. & 3 & 2 & 1 \\ \hline \end{array}$
4. $\begin{array}{|c|c|c|c|} \hline & A & B & C \\ \hline 1. & 3 & 2 & 1 \\ \hline 2. & 1 & 2 & 3 \\ \hline 3. & 1 & 3 & 2 \\ \hline 4. & 3 & 2 & 1 \\ \hline \end{array}$

Solution:

$\text{Hint: Gas has high entropy than liquid}$ $\text{A. When a liquid vaporises, i.e., liquid-vapour, entropy increases i.e., } \Delta S = \text{positive}$ $\text{B. When } \Delta H = \text{positive, i.e., energy factor opposes. The process is non-spontaneous at all temperatures if entropy also opposes the process, i.e., } \Delta S = \text{negative}$ $\text{C. In the reversible expansion of an ideal gas, the system remains in equilibrium at every stage. Hence, } \Delta S = 0$

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
}