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Current Question (ID: 15978)

Question:

$4.0 \text{ gm of gas occupies } 22.4 \text{ litres at NTP. The specific heat capacity of the gas at a constant volume is } 5.0 \text{ JK}^{-1}\text{mol}^{-1}. \text{ If the speed of sound in the gas at NTP is } 952 \text{ ms}^{-1}, \text{ then the molar heat capacity at constant pressure will be: } (R = 8.31 \text{ JK}^{-1}\text{mol}^{-1})$

Options:

1. $8.0 \text{ JK}^{-1}\text{mol}^{-1}$
2. $7.5 \text{ JK}^{-1}\text{mol}^{-1}$
3. $7.0 \text{ JK}^{-1}\text{mol}^{-1}$
4. $8.5 \text{ JK}^{-1}\text{mol}^{-1}$

Solution:

$\text{Hint: } v = \sqrt{\frac{\gamma PV}{M}}$ $\text{Step: Find the molar heat capacity at constant pressure.}$ $\text{Given: } M = 4 \text{ gm}, V = 22.4 \text{ L}, C_v = 5 \text{JK}^{-1}\text{mol}^{-1}, v_{\text{sound}} = 952 \text{m/s}, C_P = ?$ $\text{As, the velocity of sound, } v_{\text{sound}} = \sqrt{\frac{\gamma PV}{M}} \Rightarrow \gamma = \frac{Mv_{\text{sound}}^2}{PV} = \frac{C_P}{C_v}$ $\text{So, heat capacity at constant pressure, } C_P = C_v \left[ \frac{M}{PV} \right] v_{\text{sound}}^2 = 5 \left[ \frac{4 \times 10^{-3}}{10^5 \times 22.4 \times 10^{-3}} \right] (952)^2$ $= \frac{20}{22.4} \times (952)^2 \times 10^{-5}$ $= 809200 \times 10^{-5} = 8.09 \text{ J/mol K}$ $\text{Therefore, the molar heat capacity at constant pressure is } 8.09 \text{ J/mol K.}$ $\text{Hence, option (1) is the correct answer.}$

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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
}