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Current Question (ID: 19676)

Question:
$\text{The change in the magnitude of the volume of an ideal gas when a small additional pressure } \Delta P \text{ is applied at constant temperature, is the same as the change when the temperature is reduced by a small quantity } \Delta T \text{ at constant pressure. The initial temperature and pressure of the gas were } 300 \text{ K and } 2 \text{ atm. respectively. If } |\Delta T| = C|\Delta P| \text{ then the value of } C \text{ in (K/atm) is:}$
Options:
  • 1. $50$
  • 2. $100$
  • 3. $150$
  • 4. $200$
Solution:
$\text{Hint: } PV = nRT$ $P \Delta V + V \Delta P = 0 \quad (\text{for constant } T)$ $P \Delta V = nR \Delta T \quad (\text{for constant } P)$ $\Delta T = \frac{P \Delta V}{nR}$ $\Delta P = -\frac{P \Delta V}{V}$ $\left( \Delta V \text{ is same in both cases} \right)$ $\frac{\Delta T}{\Delta P} = \frac{P \Delta V}{nR} \cdot \frac{V}{-P \Delta V} = \frac{-V}{nR} = \frac{-T}{P}$ $\left( \frac{V}{nR} = \frac{T}{P} \right)$ $\left| \frac{\Delta T}{\Delta P} \right| = \left| \frac{T}{P} \right| = \frac{300}{2} = 150$

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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
}