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Current Question (ID: 19664)

Question:

$\text{Consider an ideal gas confined in an isolated closed chamber. As the gas undergoes an adiabatic expansion, the average time of collision between molecules increases as } V^q, \text{ where } V \text{ is the volume of the gas. The value of } q \text{ is: } \left( \gamma = \frac{C_P}{C_V} \right)$

Options:

1. $\frac{3\gamma + 5}{6}$
2. $\frac{3\gamma - 5}{6}$
3. $\frac{\gamma + 1}{2}$
4. $\frac{\gamma - 1}{2}$

Solution:

$\text{Hint: } TV^{\gamma - 1} = \text{constant for adiabatic process.}$ $\text{Average distance between two molecules is given by}$ $l = \lambda = \frac{1}{\sqrt{2\pi d^2 n}}$ $\therefore l = \frac{1}{\sqrt{2\pi d^2} \cdot (\text{No. of molecules})} \cdot V$ $\text{or } l \propto V$ $\text{Average time for collision is}$ $t \propto \frac{2l}{V_x} = \frac{2\sqrt{3}l}{V_{rms}}$ $t \propto \frac{V}{\sqrt{T}} \quad \cdots (2)$ $\text{For adiabatic process}$ $TV^{\gamma - 1} = \text{const.}$ $T \propto \frac{1}{V^{\gamma - 1}} \quad \cdots (3)$ $\text{From (2) and (3)}$ $t \propto V \cdot V^{\frac{\gamma - 1}{2}}$ $t \propto V^{\frac{\gamma + 1}{2}}$ $\therefore \text{On comparing with } t \propto V^q$ $\therefore q = \frac{\gamma - 1}{2}$

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{
  "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
}