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Current Question (ID: 10997)

Question:
$\text{The initial pressure and volume of a gas are } P \text{ and } V\text{, respectively. First, it is expanded isothermally to volume } 4V \text{ and then compressed adiabatically to volume } V\text{. The final pressure of the gas will be: [Given: } \gamma = 1.5\text{]}$
Options:
  • 1. $P$
  • 2. $2P$ (Correct)
  • 3. $4P$
  • 4. $8P$
Solution:
$\text{In the isothermal process, } P_1V_1 = P_2V_2$ $\text{or } PV = P_2 \times 4V$ $\therefore P_2 = \frac{P}{4}$ $\text{In the adiabatic process,}$ $P_2V_2^\gamma = P_3V_3^\gamma$ $\Rightarrow \frac{P}{4} \times (4V)^{1.5} = P_3V^{1.5}$ $\Rightarrow P_3 = 2P$

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
}