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Current Question (ID: 11058)

Question:

$\text{A sample of 0.1 g of water at 100°C and normal pressure (1.013} \times 10^5 \text{ N m}^{-2}\text{) requires 54 cal of heat energy to convert it into steam at 100°C. If the volume of the steam produced is 167.1 cc, then the change in internal energy of the sample will be:}$

Options:

1. $104.3 \text{ J}$
2. $208.7 \text{ J}$ (Correct)
3. $42.2 \text{ J}$
4. $84.5 \text{ J}$

Solution:

$\text{According to the first law of thermodynamics:}$ $\Delta Q = \Delta U + \Delta W$ $\Delta U = \Delta Q - \Delta W$ $\text{Given: Heat supplied } \Delta Q = 54 \text{ cal} = 54 \times 4.18 \text{ J} = 225.7 \text{ J}$ $\text{Work done by the system: } \Delta W = P(\Delta V)$ $\text{Initial volume of water } \approx 0.1 \text{ g} \times 1 \text{ g/cm}^3 = 0.1 \text{ cm}^3$ $\text{Final volume of steam } = 167.1 \text{ cm}^3$ $\Delta V = 167.1 - 0.1 = 167.0 \text{ cm}^3 = 167.0 \times 10^{-6} \text{ m}^3$ $\Delta W = 1.013 \times 10^5 \times 167.0 \times 10^{-6} = 16.9 \text{ J}$ $\text{Therefore:}$ $\Delta U = 225.7 - 16.9 = 208.8 \text{ J} \approx 208.7 \text{ J}$

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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
}