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Current Question (ID: 8341)

Question:

$\text{1 mole of an ideal gas at 25}^{\circ} \text{C is subjected to expand reversibly ten times of its initial volume. The change in entropy of expansion is:}$

Options:

1. $19.15 \text{ JK}^{-1}\text{mol}^{-1}$
2. $16.15 \text{ JK}^{-1}\text{mol}^{-1}$
3. $22.15 \text{ JK}^{-1}\text{mol}^{-1}$
4. $\text{None of the above}$

Solution:

$\text{Hint:}$ $\Delta\text{S} = \text{q/T}$ $\text{For the reversible isothermal expansion of an ideal gas, we can calculate the entropy change using the formula:}$ $\Delta\text{S} = \frac{2.303 \times \text{nRT}}{\text{T}} \log \frac{\text{V}_2}{\text{V}_1} = 2.303 \times \text{nR} \times \log \frac{\text{V}_2}{\text{V}_1}$ $\text{Where:}$ $\text{n = number of moles = 1 mole}$ $\text{R = universal gas constant = 8.314 J mol}^{-1}\text{ K}^{-1}$ $\text{V}_1 = \text{initial volume}$ $\text{V}_2 = \text{final volume = 10} \times \text{V}_1$ $\text{T = temperature = 25}^{\circ}\text{C = 298 K (though T cancels out in the equation)}$ $\text{Substituting these values:}$ $\Delta\text{S} = 2.303 \times 1 \times 8.314 \times \log \frac{10\text{V}_1}{\text{V}_1}$ $\Delta\text{S} = 2.303 \times 8.314 \times \log 10$ $\text{Since } \log 10 = 1:$ $\Delta\text{S} = 2.303 \times 8.314 \times 1$ $\Delta\text{S} = 19.15 \text{ JK}^{-1}\text{mol}^{-1}$ $\text{Therefore, the entropy change for the expansion is 19.15 JK}^{-1}\text{mol}^{-1}\text{, which corresponds to option 1.}$

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
}