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Current Question (ID: 19670)

Question:
$n$ \text{ moles of an ideal gas with constant volume heat capacity } $C_V$ \text{ undergo an isobaric expansion by a certain volume. The ratio of the work done in the process, to the heat supplied is:}
Options:
  • 1. $\frac{nR}{C_V + nR}$
  • 2. $\frac{nR}{C_V - nR}$
  • 3. $\frac{4nR}{C_V + nR}$
  • 4. $\frac{4nR}{C_V - nR}$
Solution:
$\text{Hint: In the isobaric process } W = nRdT$ $\frac{w}{Q} = \frac{Q - \Delta U}{nC_p dT}$ $\frac{w}{Q} = \frac{nRdT}{nC_p dT} = \frac{R}{C_p} = \frac{R}{C_v + R}$ $= \frac{nR}{nC_v + nR} = \frac{R}{C_v + R}$

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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
}