Import Question JSON

Current Question (ID: 8203)

Question:
At moderate pressure, the van der Waal's equation becomes:
Options:
  • 1. $PV_m = RT$
  • 2. $P(V_m-b) = RT$
  • 3. $(P+a/V_m^2)(V_m) = RT$
  • 4. $P = RT/V_m + a/V_m^2$
Solution:
Hint: At moderate\n\nExplanation:\n\nAs the pressure is increased, the volume decreases and hence, the factor $a/V^2$ increases. Thus, $\frac{a}{V^2}$ term cannot be neglected. If pressure is not too high, then the volume of the individual molecule is very less with respect to the volume of gas. Hence, $V-b \approx V_m$\n\nThe real gas equation at moderate temperature is:\n\n$(P+\frac{a}{V^2})(V_m) = RT$

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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
}