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Current Question (ID: 8324)

Question:

$\text{The maximum work done in expanding 16g O}_2 \text{ isothermally at 300 K and occupying a volume of}$ $\text{5 dm}^3 \text{ until the volume becomes 25 dm}^3 \text{ is:}$

Options:

1. $-2.01 \times 10^3 \text{ J}$
2. $2.01 \times 10^{-3} \text{ J}$
3. $+2.81 \times 10^3 \text{ J}$
4. $+2.01 \times 10^{-6} \text{ J}$

Solution:

$\text{Hint: work done} = -\text{nRT}\ln \frac{\text{V}_2}{\text{V}_1}$ $\text{Step 1: Identify the given values}$ $\text{The given values are as follows:}$ $\text{T = 300 K}$ $\text{V}_1 = 5 \text{ dm}^3$ $\text{V}_2 = 25 \text{ dm}^3$ $\text{Mass of O}_2 = 16 \text{ g}$ $\text{R = 8.314 J/(mol·K) (universal gas constant)}$ $\text{Step 2: Calculate the number of moles of O}_2$ $\text{Molar mass of O}_2 = 32 \text{ g/mol}$ $\text{Number of moles (n)} = \frac{\text{Mass of O}_2}{\text{Molar mass of O}_2} = \frac{16 \text{ g}}{32 \text{ g/mol}} = 0.5 \text{ mol}$ $\text{Step 3: Calculate the maximum work done during isothermal expansion}$ $\text{For an isothermal reversible process, the maximum work done is given by:}$ $\text{work done} = -\text{nRT}\ln\frac{\text{V}_2}{\text{V}_1}$ $\text{Substituting the values:}$ $\text{work done} = -(0.5 \text{ mol})(8.314 \text{ J/(mol·K)})(300 \text{ K})\ln\frac{25 \text{ dm}^3}{5 \text{ dm}^3}$ $\text{work done} = -(0.5)(8.314)(300)\ln(5)$ $\text{work done} = -1247.1 \times \ln(5)$ $\text{work done} = -1247.1 \times 1.6094$ $\text{work done} = -2007.1 \text{ J} \approx -2.01 \times 10^3 \text{ J}$ $\text{The negative sign indicates that work is done by the system on the surroundings during expansion.}$ $\text{Therefore, the maximum work done in expanding 16g of O}_2 \text{ isothermally from 5 dm}^3 \text{ to 25 dm}^3 \text{ at 300 K is } -2.01 \times 10^3 \text{ J.}$

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{
  "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
}