Import Question JSON

Current Question (ID: 19889)

Question:

$\text{The temperature of a gas is } -78^\circ \text{C and the average translational kinetic energy of its molecules is } K. \text{ The temperature at which the average translational kinetic energy of the molecules of the same gas becomes } 2K \text{ is}$

Options:

1. $-39^\circ \text{C}$
2. $127^\circ \text{C}$
3. $-78^\circ \text{C}$
4. $117^\circ \text{C}$

Solution:

$\text{The average translational kinetic energy of a gas is directly proportional to its absolute temperature.}$ $\text{Let the initial temperature in Kelvin be } T_1 = -78^\circ \text{C} + 273 = 195 \text{ K.}$ $\text{If the kinetic energy becomes } 2K, \text{ then the new temperature } T_2 \text{ should be such that } \frac{T_2}{T_1} = 2.$ $\text{Thus, } T_2 = 2 \times 195 = 390 \text{ K.}$ $\text{Converting back to Celsius: } 390 - 273 = 117^\circ \text{C.}$

Import JSON File

Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}