Import Question JSON

Current Question (ID: 11073)

Question:

$\text{An ideal gas undergoes a cyclic process ABCA as shown. The heat exchange between the system and the surrounding during the process will be:}$

Options:

1. $10 \text{ J}$
2. $5 \text{ J}$ (Correct)
3. $15 \text{ J}$
4. $20 \text{ J}$

Solution:

$\text{The heat exchange between the system and the surroundings during the cyclic process ABCA is 5 J.}$ $\text{The net heat exchange for a cyclic process is equal to the net work done by the system. The net work is the area enclosed by the cycle on the pressure-volume (P-V) diagram.}$ $\text{In this case, the cycle is a triangle. The area of the triangle can be calculated using its base and height.}$ $\text{Base: The difference in volume is } 4 \text{ m}^3 - 2 \text{ m}^3 = 2 \text{ m}^3$ $\text{Height: The difference in pressure is } 10 \text{ N/m}^2 - 5 \text{ N/m}^2 = 5 \text{ N/m}^2$ $\text{The area is given by the formula:}$ $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$ $\text{Area} = \frac{1}{2} \times 2 \text{ m}^3 \times 5 \text{ N/m}^2 = 5 \text{ J}$ $\text{Since the cycle is traced in a clockwise direction, the work done by the gas is positive. Therefore, the heat exchange between the system and the surroundings is 5 J.}$

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
}