Import Question JSON

Current Question (ID: 11020)

Question:

$\text{The pressure of a monoatomic gas increases linearly from } 4 \times 10^5 \text{ N/m}^2 \text{ to } 8 \times 10^5 \text{ N/m}^2 \text{ when its volume increases from } 0.2 \text{ m}^3 \text{ to } 0.5 \text{ m}^3. \text{ The work done by the gas is:}$

Options:

1. $2.8 \times 10^5 \text{ J}$
2. $1.8 \times 10^6 \text{ J}$
3. $1.8 \times 10^5 \text{ J}$ (Correct)
4. $1.8 \times 10^2 \text{ J}$

Solution:

$\text{Hint: } \Delta W = \int P dV = \text{The area under P-V curve}$ $\text{Step: Find the work done by the gas.}$ $\text{Since the pressure increases linearly with volume, the P-V diagram shows a trapezoid.}$ $\text{Given:}$ $\text{Initial pressure } P_i = 4 \times 10^5 \text{ N/m}^2$ $\text{Final pressure } P_f = 8 \times 10^5 \text{ N/m}^2$ $\text{Initial volume } V_i = 0.2 \text{ m}^3$ $\text{Final volume } V_f = 0.5 \text{ m}^3$ $\text{The work done by the gas is given by:}$ $\Delta W = \int P dV = \text{Area under P-V curve}$ $\text{For a linear pressure change, the area is that of a trapezoid:}$ $\Delta W = \frac{1}{2}[V_f - V_i] \times [P_f + P_i]$ $\Delta W = \frac{1}{2}[0.5 - 0.2] \times [8 \times 10^5 + 4 \times 10^5]$ $\Delta W = \frac{1}{2} \times 0.3 \times 12 \times 10^5$ $\Delta W = 1.8 \times 10^5 \text{ J}$ $\text{Hence, option (3) is the correct answer.}$

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
}