Import Question JSON

Current Question (ID: 11115)

Question:

$\text{Which one, of the following, graphs represents the behaviour of an ideal gas at constant temperature?}$

Options:

1. $\text{Graph representing a linear increase of } \text{PV} \text{ with } \text{V}.$
2. $\text{Graph representing a constant value of } \text{PV} \text{ with } \text{V}.$ (Correct)
3. $\text{Graph representing a linear decrease of } \text{PV} \text{ with } \text{V}.$
4. $\text{Graph representing an inverse relationship of } \text{PV} \text{ with } \text{V}.$

Solution:

$\text{Hint: At constant temperature, the product } \text{PV} \text{ remains constant.}$ \\ $\text{Step 1: Recall the equation for an isothermal process. According to Boyle's law, for a fixed amount of an ideal gas at constant temperature, the pressure and volume are inversely proportional.}$ \\ $\text{PV} = \text{constant}$ \\ $\text{This means that as the volume } \text{V} \text{ changes, the product } \text{PV} \text{ remains a constant value.}$ \\ $\text{Step 2: The product } \text{PV} \text{ is independent of } \text{V} \text{. Therefore, the graph of } \text{PV} \text{ versus } \text{V} \text{ will be a horizontal line, representing a constant value.}$ \\ $\text{The correct graph is the one that shows a horizontal line for the product of pressure and volume } (\text{PV}) \text{ as a function of volume } (\text{V}) \text{.}$

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
}