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Current Question (ID: 8194)

Question:

$\text{The kinetic energy for 14 }\textit{grams}\text{ of nitrogen gas at }127^{\circ}\text{C is nearly - }$\n\n$(\text{mol. mass of nitrogen = 28 and gas constant = 8.31JK}^{-1}\text{mol}^{-1})$

Options:

1. $1.0 \text{ J}$
2. $4.15 \text{ J}$
3. $2493 \text{ J}$
4. $3.3 \text{ J}$

Solution:

$\text{HINT: Use a general expression for kinetic energy (K.E. }=3/2\text{RT)}$\n\n$\text{Step 1:}$\n\n$\text{The formula of Kinetic energy for } \textbf{one molecule} \text{ is as follows:}$\n\n$\text{K.E} = \frac{3}{2}kT$\n\n$\text{here, k is Boltzmann constant.}$\n\n$\text{The relation between R (gas constant) and k is as follows:}$\n\n$\text{R=N}_A\text{K}$\n\n$\text{N}_A \text{ is a Avogadro constant.}$\n\n$\text{For one molecule the relation is R = k}$\n\n$\text{Hence, the formula of kinetic energy for one molecule with respect to R is:}$\n\n$\text{K.E} = \frac{3}{2}RT$\n\n$\text{For N}_A \text{ molecule it is K.E} = \frac{3}{2}N_A RT$\n\n$\text{1 mole = N}_A$\n\n$\text{For } \textbf{one mole} \text{ the formula is K.E} = \frac{3}{2}RT$\n\n$\text{The formula of kinetic energy for n mole of gas is as follows:}$\n\n$\text{K.E} = \frac{3}{2}nRT$\n\n$\text{Step 2:}$\n\n$\text{Calculate the value of kinetic energy for n mole is as follows:}$\n\n$\text{or K.E.} = \frac{3}{2}\text{nRT}$\n\n$\text{K.E} = \frac{3}{2} \times \frac{14}{28} \times 8.31 \times 400 \text{ J} = 2493 \text{ J}$

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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
}