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Current Question (ID: 11060)

Question:
$\text{The figure below shows two paths that may be taken by a gas to go from state A to state C. In process AB, 400 J of heat is added to the system and in process BC, 100 J of heat is added to the system. The heat absorbed by the system in the process AC will be:}$ $\text{Given P-V diagram with coordinates: A}(2 \times 10^{-3} \text{ m}^3, 2 \times 10^4 \text{ Pa})\text{, B}(2 \times 10^{-3} \text{ m}^3, 6 \times 10^4 \text{ Pa})\text{, C}(4 \times 10^{-3} \text{ m}^3, 6 \times 10^4 \text{ Pa})$
Options:
  • 1. $380 \text{ J}$
  • 2. $500 \text{ J}$
  • 3. $460 \text{ J}$ (Correct)
  • 4. $300 \text{ J}$
Solution:
$\text{Since the initial and final points are the same:}$ $\Delta U_{A \rightarrow B \rightarrow C} = \Delta U_{A \rightarrow C} \text{ ......(i)}$ $\text{Also A } \rightarrow \text{ B is isochoric process}$ $\text{So } \Delta W_{A \rightarrow B} = 0$ $\text{and } \Delta Q = \Delta U + \Delta W$ $\text{So, } \Delta Q_{A \rightarrow B} = \Delta U_{A \rightarrow B} = 400 \text{ J}$ $\text{Next B } \rightarrow \text{ C is isobaric process.}$ $\text{So, } \Delta Q_{B \rightarrow C} = \Delta U_{B \rightarrow C} + \Delta W_{B \rightarrow C}$ $= \Delta U_{B \rightarrow C} + P \Delta V_{B \rightarrow C}$ $\Rightarrow 100 = \Delta U_{B \rightarrow C} + 6 \times 10^4 (2 \times 10^{-3})$ $\Rightarrow \Delta U_{B \rightarrow C} = 100 - 120 = -20 \text{ J}$ $\text{From Equation (i):}$ $\Delta U_{A \rightarrow B} + \Delta U_{B \rightarrow C} = \Delta Q_{A \rightarrow C} - \Delta W_{A \rightarrow C}$ $\Rightarrow 400 + (-20) = \Delta Q_{A \rightarrow C} - \text{Area under AC}$ $\Delta Q_{A \rightarrow C} = 380 + (2 \times 10^4 \times 2 \times 10^{-3} + \frac{1}{2} \times 2 \times 10^{-3} \times 4 \times 10^4)$ $= 380 + (40 + 40)$ $\Delta Q_{A \rightarrow C} = 460 \text{ J}$

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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
}