Import Question JSON

Current Question (ID: 21038)

Question:

$\text{Assuming that water vapor is an ideal gas, the internal energy change } (\Delta U) \text{ when 1 mol of water is vaporized at 1 bar pressure and } 100^\circ\text{C, will be:}$ $\text{(Given: Molar enthalpy of vaporization of water at 1 bar and } 373\, \text{K} = 41\, \text{kJ mol}^{-1} \text{ and } R = 8.3\, \text{J mol}^{-1} \text{K}^{-1})$

Options:

1. $4.100\, \text{kJ mol}^{-1}$
2. $3.7904\, \text{kJ mol}^{-1}$
3. $37.904\, \text{kJ mol}^{-1}$
4. $41.00\, \text{kJ mol}^{-1}$

Solution:

$\text{Hint: } \Delta H = \Delta U + \Delta n_g RT$ $\text{Step 1: It is given that,}$ $\text{Molar enthalpy of vaporization of water} = 41\, \text{kJ mol}^{-1}$ $\text{Temperature} = 373\, \text{K}$ $\text{Universal gas constant (R)} = 8.3\, \text{J mol}^{-1} \text{K}^{-1}$ $\text{This question aims to calculate the internal energy change.}$ $\text{Step 2: } \Delta H = \Delta U + \Delta n_g RT$ $\Delta n_g = 1$ $\Delta U = \Delta H - \Delta n_g RT$ $\text{or, } \Delta U = 41\, \text{kJ mol}^{-1} - (1 \times 8.3\, \text{J mol}^{-1} \text{K}^{-1} \times 373\, \text{K})$ $\Delta U = 41\, \text{kJ mol}^{-1} - (3095.9\, \text{J mol}^{-1}) = 41\, \text{kJ mol}^{-1} - (3.0959\, \text{kJ mol}^{-1})$ $\Delta U = 37.904\, \text{kJ mol}^{-1}$ $\text{Option 3 is the correct choice.}$

Import JSON File

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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
}