Import Question JSON

Current Question (ID: 11124)

Question:

$\text{An experiment is carried out on a fixed amount of gas at different temperatures and at high pressure such that it deviates from the ideal gas behaviour. The variation of } \frac{PV}{RT} \text{ with } P \text{ is shown in the diagram. The correct variation will correspond to: (Assuming that the gas in consideration is nitrogen)}$

Options:

1. $\text{Curve A}$
2. $\text{Curve B}$ (Correct)
3. $\text{Curve C}$
4. $\text{Curve D}$

Solution:

$\text{For an ideal gas, } \frac{PV}{RT} = 1 \text{ regardless of pressure. This is represented by curve D. However, the question states that the gas deviates from ideal behaviour, especially at high pressure. At lower pressures, the given gas behaves as an ideal gas, so } \frac{PV}{RT} = \text{constant} \approx 1. \text{ As pressure increases, the gas molecules are forced closer together. The volume of the molecules themselves becomes significant and the intermolecular forces become more prominent. For nitrogen (a real gas), at high pressures, the volume occupied by the gas molecules themselves is not negligible compared to the total volume, so the measured volume is higher than expected. This causes the value of } \frac{PV}{RT} \text{ to increase with increasing pressure. Therefore, Curve B, which starts at } \frac{PV}{RT} = 1 \text{ at low pressure and increases as pressure rises, correctly represents this behavior.}$

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
}