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Current Question (ID: 11041)

Question:

$\text{Two cylinders contain the same amount of an ideal monatomic gas. The same amount of heat is given to two cylinders. If the temperature rise in cylinder A is } T_0\text{, then the temperature rise in cylinder B will be:}$ $\text{(Cylinder A has a free piston, Cylinder B has a fixed piston)}$

Options:

1. $\frac{4}{3}T_0$
2. $2T_0$
3. $\frac{T_0}{2}$
4. $\frac{5}{3}T_0$ (Correct)

Solution:

$\text{Hint: Process in cylinder A is isobaric.}$ $\text{Step 1: Find heat given to cylinder A.}$ $Q = nC_p \Delta T$ $Q = nC_p T_0 \ldots (1)$ $\text{Step 2: Find the change in internal energy in cylinder B.}$ $Q = \Delta U = nC_v \Delta T \ldots (2)$ $\text{Step 3: Equate the heat in both cases.}$ $nC_p T_0 = nC_v \Delta T$ $\Delta T = \frac{C_p}{C_v} T_0$ $= \gamma T_0$ $= \frac{5}{3} T_0$ $\text{For a monatomic ideal gas, } \gamma = \frac{C_p}{C_v} = \frac{5/2 R}{3/2 R} = \frac{5}{3}\text{. Therefore, the temperature rise in cylinder B is } \frac{5}{3}T_0\text{.}$

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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
}