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Current Question (ID: 11051)

Question:
$\text{A system is taken from state } A \text{ to state } B \text{ along two different paths, 1 and 2. If the heat absorbed and work done by the system along these two paths are } Q_1, Q_2 \text{ and } W_1, W_2 \text{ respectively, then:}$
Options:
  • 1. $Q_1 = Q_2$
  • 2. $W_1 = W_2$
  • 3. $Q_1 - W_1 = Q_2 - W_2$ (Correct)
  • 4. $Q_1 + W_1 = Q_2 + W_2$
Solution:
$\text{Hint: } \Delta Q = \Delta U + \Delta W$ $\text{Explanation: Internal energy is a state function i.e. not dependent on the path. From the first law of thermodynamics i.e., } \Delta U = 0.$ $U_A = U_B$ $\Rightarrow Q_1 - W_1 = Q_2 - W_2 \quad [\Delta U = \Delta Q - \Delta W]$ $\text{Hence, option (3) is the correct answer.}$

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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
}