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Current Question (ID: 8148)

Question:

$\text{The temperature of 4.0 mole of a gas occupying 5 dm}^3 ext{ at 3.32 bar is - (R = 0.083 bar dm}^3 ext{ K}^{-1} ext{ mol}^{-1} ext{)}.$

Options:

1. $40\text{ K}$
2. $30\text{ K}$
3. $20\text{ K}$
4. $50\text{ K}$

Solution:

$\text{Hint: Use the Ideal gas equation.} ext{Step 1: Write down the given data} ext{Given,} $ ext{n = 4.0 mol; V = 5 dm}^3 ext{; P = 3.32 bar}$ $ ext{R = 0.083 bar dm}^3 ext{ K}^{-1} ext{ mol}^{-1} ext{ and T = ?}$ $ ext{Step 2: Calculate T using the Ideal gas equation}$ $ ext{We know that, PV = nRT}$ $ ext{T} = rac{ ext{PV}}{ ext{nR}} = rac{3.32 imes 5}{4 imes 0.083} = 50 ext{ K}$ $ ext{Hence, the temperature of the gas is 50 K.}$

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
}