Import Question JSON

Current Question (ID: 8337)

Question:

$\text{The entropy change in the isothermal reversible expansion of 2 moles of an ideal gas from 10 to 100 L at 300 K is}$

Options:

1. $42.3 \text{ J K}^{-1}$
2. $35.8 \text{ J K}^{-1}$
3. $38.3 \text{ J K}^{-1}$
4. $32.3 \text{ J K}^{-1}$

Solution:

$\text{Hint:}$ $\Delta\text{S} = 2.303 \text{ nR}\log \frac{\text{V}_2}{\text{V}_1}$ $\text{For the isothermal reversible expansion of an ideal gas, we can calculate the entropy change using the formula:}$ $\Delta\text{S} = 2.303 \text{ nR}\log \frac{\text{V}_2}{\text{V}_1}$ $\text{Where:}$ $\text{n = number of moles = 2 moles}$ $\text{R = universal gas constant = 8.314 J mol}^{-1}\text{ K}^{-1}$ $\text{V}_1 = \text{initial volume = 10 L}$ $\text{V}_2 = \text{final volume = 100 L}$ $\text{Substituting these values:}$ $\Delta\text{S} = 2.303 \times \text{nR} \times \log \frac{\text{V}_2}{\text{V}_1}$ $\Delta\text{S} = 2.303 \times 2 \times 8.314 \times \log \frac{100}{10}$ $\Delta\text{S} = 2.303 \times 2 \times 8.314 \times \log 10$ $\Delta\text{S} = 2.303 \times 2 \times 8.314 \times 1$ $\Delta\text{S} = 2.303 \times 16.628$ $\Delta\text{S} = 38.3 \text{ J K}^{-1}$ $\text{Note that the final unit is J K}^{-1}\text{ (not J mol}^{-1}\text{ K}^{-1}\text{) because we have already accounted for the number of moles in our calculation.}$ $\text{Therefore, the entropy change in the isothermal reversible expansion is 38.3 J K}^{-1}\text{, which corresponds to option 3.}$

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
}