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Current Question (ID: 19857)

Question:
$\text{In a mixture, } 0.5 \text{ moles of } \text{O}_2 \text{ and } 4 \text{ moles of } \text{Ne} \text{ gas are taken at temperature } T. \text{ The internal energy of the system is equal to:}$
Options:
  • 1. $\left( \frac{13}{2} \right) RT$
  • 2. $\left( \frac{11}{4} \right) RT$
  • 3. $\left( \frac{29}{4} \right) RT$
  • 4. $\left( \frac{13}{4} \right) RT$
Solution:
$\text{Hint: } U_{\text{final}} = U_{\text{O}_2} + U_{\text{Ne}}$ $\text{Step: Find the internal energy of the system.}$ $\text{The internal energy of the system is given as:}$ $U = \frac{f_1}{2} n_1 RT + \frac{f_2}{2} n_2 RT \text{ where, } f_1 \text{ and } f_2 \text{ are degrees of freedom of } \text{O}_2 \text{ and } \text{Ne} \text{ gas respectively.}$ $n_1 \text{ and } n_2 \text{ are the numbers of moles of } \text{O}_2 \text{ and } \text{Ne} \text{ gas respectively.}$ $U = \frac{5}{2} \times \frac{1}{2} RT + \frac{3}{2} \times 4RT$ $U = \frac{5}{4} RT + 6RT = \frac{(5+24)}{4} RT = \frac{29}{4} RT$ $U = \frac{29}{4} RT$ $\text{Hence, option (3) is the correct answer.}$

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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
}