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Current Question (ID: 19663)

Question:

$\text{Consider a spherical shell of radius } R \text{ at temperature } T. \text{ The black body radiation inside it can be considered as an ideal gas of photons}$ $\text{with internal energy per unit volume } u = \frac{U}{V} \propto T^4 \text{ and } P = \frac{1}{3} \left( \frac{U}{V} \right).$ $\text{If the shell now undergoes an adiabatic expansion the relation between } T \text{ and } R \text{ is:}$

Options:

1. $T \propto e^{-R}$
2. $T \propto e^{-3R}$
3. $T \propto \frac{1}{R}$
4. $T \propto \frac{1}{R^3}$

Solution:

$\text{Hint: } dQ = 0 \text{ for adiabatic process.}$ $\text{Energy per unit volume } \frac{U}{V} \propto T^4$ $\text{The process is adiabatic so } dQ = 0$ $dQ = du + dw \text{ (from 1st law of thermodynamics)}$ $dQ = 0$ $du + dw = 0$ $dw = -du$ $PdV = -du$ $\text{given } P = \frac{1}{3} \frac{U}{V}$ $\frac{1}{3} \frac{U}{V} dV = -du$ $\Rightarrow \frac{du}{U} + \frac{1}{3} \frac{dV}{V} = 0$ $\text{on integrating}$ $\ln U + \frac{1}{3} \ln V = \text{const.}$ $\ln (UV^{1/3}) = \text{const.}$ $\Rightarrow UV^{1/3} = \text{const.} \quad \ldots (i)$ $\text{given } \frac{U}{V} \propto T^4$ $\text{So } U \propto VT^4$ $\text{Substitute in equation (1)}$ $VT^4V^{1/3} = \text{const}$ $T^4V^{4/3} = \text{const}$ $TV^{1/3} = \text{const}$ $V = \frac{4}{3} \pi R^3$ $\text{So } T \left( \frac{4}{3} \pi R^3 \right)^{1/3} = \text{const}$ $TR = \text{const.}$ $\Rightarrow T \propto \frac{1}{R}$

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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
}