Import Question JSON

Current Question (ID: 8127)

Question:
$\text{The plots of different values of pressure versus temperature are shown in the given figure. The correct order of volume for the following plot is-}$
Options:
  • 1. $V_4 < V_3 < V_2 < V_1$
  • 2. $V_4 > V_3 < V_2 > V_1$
  • 3. $V_4 > V_3 > V_2 > V_1$
  • 4. $V_4 > V_3 > V_2 < V_1$
Solution:
$\text{Gay Lussac's law states that at constant volume, the pressure of a fixed amount of a gas varies directly with the temperature.}$ $P \propto T$ $\frac{P}{T} = \text{Constant}$ $\text{From the ideal gas equation, } PV = nRT \text{, we can write } V = \frac{nRT}{P}\text{.}$ $\text{For a fixed temperature (e.g., draw a vertical line on the graph), as we move from } V_1 \text{ to } V_4 \text{, the pressure decreases.}$ $\text{Since } V = \frac{nRT}{P} \text{, at constant temperature, if pressure (P) decreases, volume (V) must increase.}$ $\text{Therefore, for a given temperature, } P_1 > P_2 > P_3 > P_4 \text{ corresponds to } V_1 < V_2 < V_3 < V_4 \text{.}$ $\text{Alternatively, consider a constant pressure (horizontal line). As we move from } V_1 \text{ to } V_4 \text{, the temperature increases.}$ $\text{Since } V \propto T \text{ at constant pressure (Charles' Law), if temperature increases, volume increases.}$ $\text{Thus, for a fixed pressure, since } T_1 < T_2 < T_3 < T_4 \text{ (implied by the slopes as } P/T \text{ is related to } 1/V \text{), it means } V_1 < V_2 < V_3 < V_4 \text{.}$ $\text{Here, option third is the correct answer.}$

Import JSON File

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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
}