Import Question JSON

Current Question (ID: 10992)

Question:

$\text{If an ideal gas undergoes two processes at constant volumes } V_1 \text{ and } V_2 \text{ as shown in the pressure-temperature } (P-T) \text{ diagram, then:}$

Options:

1. $V_1 = V_2$
2. $V_1 > V_2$
3. $V_1 < V_2$ (Correct)
4. $V_1 \geq V_2$

Solution:

$\text{Based on the pressure-temperature (P-T) diagram, the correct relationship between the two constant volumes is } V_1 < V_2\text{.}$ $\text{This conclusion is derived from the ideal gas law, } PV = nRT\text{, which can be rearranged to express pressure as a function of temperature:}$ $P = \left(\frac{nR}{V}\right)T$ $\text{In this equation, the term } \left(\frac{nR}{V}\right) \text{ represents the slope of the line in the P-T diagram.}$ $\text{From the graph, the line for } V_1 \text{ is steeper than the line for } V_2\text{, which means:}$ $\text{Slope of } V_1 > \text{Slope of } V_2$ $\frac{nR}{V_1} > \frac{nR}{V_2}$ $\text{Since the number of moles } (n) \text{ and the ideal gas constant } (R) \text{ are positive and constant, the inequality simplifies to:}$ $\frac{1}{V_1} > \frac{1}{V_2}$ $\text{Taking the reciprocal of both sides reverses the inequality sign, leading to the final relationship:}$ $V_1 < V_2$

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
}