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Current Question (ID: 8202)

Question:

For one mole of van der Waal's gas when $b=0$ and $T=300$ K, the PV vs. $1/V$ plot is shown below. The value of the van der Waal's constant 'a' ($\text{atm. litre}^2 \text{mol}^{-2}$) is-

Options:

1. $1.0$
2. $4.5$
3. $1.5$
4. $3.0$

Solution:

Hint: $\left(P + \frac{a}{V^2}\right)(V - b) = RT$\n\nStep 1:\nThe value of b is zero. Hence, the van der waals equation is as follows:\n\n$\left(P + \frac{a}{V^2}\right)V = RT$\n\nStep 2:\n\nCalculate the value of a as follows:\n\n$PV + \frac{a}{V} = RT$\n\n$PV = RT - \frac{a}{V}$\n\nSo, plot of $(PV)$ and $\left(\frac{1}{V}\right)$ has slope $-a$.\n\n$\therefore \text{slope} = \frac{20.1-24.6}{3} = -1.5$\n\n$\therefore a = 1.5 \text{ atm L}^2 \text{mol}^{-2}$

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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
}