Import Question JSON

Current Question (ID: 8267)

Question:

$\text{As an isolated box, equally partitioned, contains two ideal gasses A and B as shown:}$ $\text{[Box diagram showing two compartments: left compartment contains gas A at 1 atm, 25°C, right compartment contains gas B at 1 atm, 25°C]}$ $\text{When the partition is removed, the gases mix. The changes in enthalpy (}\Delta\text{H) and entropy (}\Delta\text{S) in the process, respectively, are}$

Options:

1. $\text{Zero, positive}$
2. $\text{Zero, negative}$
3. $\text{Positive, zero}$
4. $\text{Negative, zero}$

Solution:

$\text{Hint: Enthalpy remains constant}$ $\text{According to KTG (Kinetic Theory of Gases)}$ $\text{Force of attraction and repulsion amongst molecules of ideal gas are negligible}$ $\text{So, } \Delta\text{H} = 0$ $\text{and randomness increases due to increase in volume so } \Delta\text{S} = \text{ + ve}$ $\text{For ideal gases, when the partition is removed, there is no energy change because ideal gas molecules do not interact with each other. Therefore, the enthalpy change (}\Delta\text{H) is zero. However, the gases now have more volume to occupy, which increases the number of possible molecular arrangements. This increased randomness means the entropy change (}\Delta\text{S) is positive.}$

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
}