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Current Question (ID: 11168)

Question:

$\text{The ratio of the specific heats } \frac{C_P}{C_V} = \gamma \text{ in terms of degrees of freedom } (n) \text{ is given by:}$

Options:

1. $1 + \frac{1}{n}$
2. $1 + \frac{n}{3}$
3. $1 + \frac{2}{n}$ (Correct)
4. $1 + \frac{n}{2}$

Solution:

$\text{Hint: } C_P - C_V = R$ $\text{Step: Find the ratio of the specific heats.}$ $\text{The internal energy for 1 mole of gas is given as } U = \frac{n}{2} RdT = C_V dT$ $\text{where } n \text{ is degrees of freedom}$ $C_P - C_V = R$ $C_P = \left(1 + \frac{n}{2}\right) R$ $\text{and } \frac{C_P}{C_V} = \gamma = \frac{\left(1+\frac{n}{2}\right) R}{\frac{n}{2} R} = 1 + \frac{2}{n}$ $\text{Hence, option (3) is the correct answer.}$

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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
}