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Current Question (ID: 11023)

Question:
$\text{An ideal gas is taken from point A to point B, as shown in the } P\text{-}V \text{ diagram. The work done in the process is:}$
Options:
  • 1. $(P_A - P_B)(V_B - V_A)$
  • 2. $\frac{1}{2}(P_B - P_A)(V_B + V_A)$
  • 3. $\frac{1}{2}(P_B - P_A)(V_B - V_A)$
  • 4. $\frac{1}{2}(P_B + P_A)(V_B - V_A)$ (Correct)
Solution:
$\text{Hint: } W = \text{Area bounded by the indicator diagram with } V\text{-axis}$ $\text{Solution:}$ $\text{The work done by a gas in a thermodynamic process is equal to the area under the } P\text{-}V \text{ curve.}$ $\text{For the linear process from point A to point B shown in the diagram, the area forms a trapezoid.}$ $\text{The area of a trapezoid is: Area} = \frac{1}{2}(\text{sum of parallel sides}) \times (\text{height})$ $\text{In this case, the parallel sides are the pressures } P_A \text{ and } P_B\text{, and the height is } (V_B - V_A)$ $\text{Therefore: } W = \frac{1}{2}(P_A + P_B)(V_B - V_A)$

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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
}