Import Question JSON

Current Question (ID: 8281)

Question:

$\text{The standard enthalpy of vaporisation } \Delta_{\text{vap}}H^{\circ} \text{ for water at } 100^{\circ} \text{C is } 40.66 \text{ kJ mol}^{-1}\text{. The internal energy of vaporisation of water at } 100^{\circ} \text{C (in kJ mol}^{-1}\text{) is-}$ $\text{(Assume water vapour to behave like an ideal gas)}$

Options:

1. $+37.56$
2. $-43.76$
3. $+43.76$
4. $+40.66$

Solution:

$\text{Hint: } \Delta_{\text{vap}}H^{\circ} = \Delta_{\text{vap}}E^{\circ} + \Delta n_g RT$ $\text{Step 1:}$ $\text{In this reaction, the vaporisation of water takes place. The reaction is as follows:}$ $\text{H}_2\text{O}_{(l)} \xrightarrow{100^{\circ} \text{C}} \text{H}_2\text{O}_{(g)}$ $\text{The relation between } \Delta_{\text{vap}}H^{\circ} \text{ and } \Delta_{\text{vap}}E^{\circ} \text{ is as follows:}$ $\Delta_{\text{vap}}H^{\circ} = \Delta_{\text{vap}}E^{\circ} + \Delta n_g RT$ $\text{Calculate the value of } \Delta n_g \text{ as follows:}$ $\Delta n_g = n_p - n_r = 1 - 0 = 1$ $\text{Step 2:}$ $\text{Calculate the value of } \Delta_{\text{vap}}E^{\circ} \text{ is as follows:}$ $\therefore 40.66 \text{ kJ mol}^{-1} = \Delta_{\text{vap}}E^{\circ} + 1 \times 8.314 \times 10^{-3} \times 373$ $\Delta_{\text{vap}}E^{\circ} = 40.66 \text{ kJ mol}^{-1} - 3.1 \text{ kJ mol}^{-1}$ $= +37.56 \text{ kJ mol}^{-1}$ $\text{Hence, option first is the answer.}$

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
}