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Current Question (ID: 11059)

Question:

$\text{In an adiabatic expansion of a gas, if the initial and final temperatures are } T_1 \text{ and } T_2\text{, respectively, then the change in internal energy of the gas is:}$

Options:

1. $\frac{nR}{\gamma-1} (T_2 - T_1)$ (Correct)
2. $\frac{nR}{\gamma-1} (T_1 - T_2)$
3. $nR (T_1 - T_2)$
4. $\text{Zero}$

Solution:

$\text{For an adiabatic process, } Q = 0 \text{ (no heat exchange)}$ $\text{From the first law of thermodynamics: } \Delta U = Q - W$ $\text{Since } Q = 0\text{, we have } \Delta U = -W$ $\text{For an adiabatic process, the work done is:}$ $W = \frac{nR(T_1 - T_2)}{\gamma - 1}$ $\text{Therefore:}$ $\Delta U = -W = -\frac{nR(T_1 - T_2)}{\gamma - 1} = \frac{nR(T_2 - T_1)}{\gamma - 1}$

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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
}