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Current Question (ID: 11111)

Question:

$\text{The volume } V \text{ versus temperature } T \text{ graph for a certain amount of a perfect gas at two pressures } P_1 \text{ and } P_2 \text{ are shown in the figure. Here:}$

Options:

1. $P_1 < P_2$
2. $P_1 > P_2$ (Correct)
3. $P_1 = P_2$
4. $\text{Pressures can't be related}$

Solution:

$\text{Hint: } \frac{PV}{T} = \text{constant. Step: Find the relation of a perfect gas at two pressures } P_1 \text{ and } P_2 \text{. The ideal gas equation is given by; } PV = nRT \text{. } V = \frac{nR}{P} \times T \text{. Thus in } V-T \text{ curve; This represents a straight line with having slope } m = \frac{nR}{P} \text{. For the same amount of gas; Slope } \propto \frac{1}{P} \text{. } P_1 > P_2 \text{. At high pressure, the slope of the straight line is less. Hence, option (2) is the correct answer.}$

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Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}