Import Question JSON

Current Question (ID: 8379)

Question:

$\text{The volume of the container containing a liquid and its vapours at a constant temperature is suddenly increased. What would be the effect of the change on vapour pressure?}$

Options:

1. $\text{It would decrease initially.}$
2. $\text{It would increase initially.}$
3. $\text{It would remain the same.}$
4. $\text{None of the above}$

Solution:

$\text{Hint: } P \propto \frac{1}{V}$ $\text{Explanation:}$ $\text{If the volume of the container is suddenly increased, then the vapour pressure would decrease initially. This is because the amount of vapour remains the same, but the volume increases suddenly. As a result, the same amount of vapour is distributed in a larger volume.}$ $\text{According to the ideal gas law: } PV = nRT$ $\text{At constant temperature (T) and constant amount of vapour (n), pressure is inversely proportional to volume:}$ $P \propto \frac{1}{V}$ $\text{When volume increases suddenly:}$ $\text{• The number of vapour molecules remains constant initially}$ $\text{• These molecules are now distributed in a larger volume}$ $\text{• This results in lower concentration of vapour molecules}$ $\text{• Therefore, vapour pressure decreases initially}$ $\text{Note: Eventually, more liquid will evaporate to restore equilibrium vapour pressure at that temperature.}$

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
}