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Current Question (ID: 8208)

Question:

The van der Waals' equation for real gas is:

Options:

1. $(P + a/V^2)(V-b) = RT$
2. $(P+n^2a/V^2)(V-nb) = nRT$
3. $P = \frac{nRT}{V-nb} - \frac{an^2}{V^2}$
4. All of the above

Solution:

Hint: $(P + a/V^2)(V-b) = RT$\n\nThe Van der Waal's equation for n mole of gas is as follows:\n\n$(P+\frac{n^2a}{V^2})(V-nb) = nRT$\n\nThe Van der Waal equation of 1 mole of gas is as follows:\n\n$(P+\frac{a}{V^2})(V-b) = RT$\n\n$P=nRT/(V-nb) - (an^2/V^2)$\n\nThese are van der Waals' equations for 1 mole (a) and n mole gas (b), (c).

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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
}