Import Question JSON

Current Question (ID: 11139)

Question:

$\text{Diatomic molecules like hydrogen have energies due to both translational as well as rotational motion. The equation in kinetic theory } \text{PV} = \frac{2}{3} \text{E}, \text{E is:}$

Options:

1. $\text{the total energy per unit volume.}$
2. $\text{only the translational part of energy because rotational energy is very small compared to translational energy.}$
3. $\text{only the translational part of the energy because during collisions with the wall, pressure relates to change in linear momentum.}$ (Correct)
4. $\text{the translational part of the energy because rotational energies of molecules can be of either sign and its average over all the molecules is zero.}$

Solution:

$\text{According to the kinetic theory of gases, the pressure exerted by a gas is a result of the transfer of linear momentum when gas molecules collide with the container walls. The equation } \text{PV} = \frac{2}{3} \text{E} \text{ is derived from this principle. In this derivation, only the change in the linear momentum of the molecules in the direction perpendicular to the wall is considered. Rotational motion and energy do not contribute to this change in linear momentum, as they do not affect the translational velocity components that cause collisions with the walls. Therefore, the term E in the equation represents only the translational kinetic energy of the gas molecules. This is because the pressure is a direct result of the translational motion of the molecules. Thus, option 3 is the correct answer.}$

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
}