Import Question JSON

Current Question (ID: 11141)

Question:

$\text{1 mole of an ideal gas is contained in a cubical volume V, ABCDEFGH at 300 K (figure). One face of the cube (EFGH) is made up of a material which totally absorbs any gas molecule incident on it. At any given time:}$

Options:

1. $\text{the pressure on EFGH would be zero.}$
2. $\text{the pressure on all the faces will be equal.}$
3. $\text{the pressure on EFGH would be double the pressure on ABCD.}$
4. $\text{the pressure on EFGH would be half that on ABCD.}$ (Correct)

Solution:

$\text{The pressure exerted by a gas is a result of the change in momentum of the gas molecules as they collide with the walls of the container. For an ideal gas molecule colliding elastically with a wall, the change in momentum is } \Delta\text{p} = 2\text{mv}$, $\text{where m is the mass and v is the velocity of the molecule. The wall ABCD is a normal wall, and the molecules collide elastically with it. Therefore, the pressure on face ABCD is proportional to the change in momentum, } \text{P}_{\text{ABCD}} \propto 2\text{mv}$. $\text{For the face EFGH, which absorbs the molecules, there is no rebound. The molecule stops, and the change in momentum is only } \Delta\text{p} = \text{mv}$. $\text{As pressure is directly proportional to the momentum transferred to the wall per unit time, the pressure on face EFGH will be proportional to this change in momentum, } \text{P}_{\text{EFGH}} \propto \text{mv}$. $\text{By comparing the two pressures, it is clear that } \text{P}_{\text{EFGH}} = \frac{1}{2} \text{P}_{\text{ABCD}}$. $\text{Therefore, the pressure on EFGH would be half that on ABCD. This corresponds to option 4.}$

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
}