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Current Question (ID: 8286)

Question:

$\text{Two moles of an ideal gas is heated at a constant pressure of one atmosphere from } 27^{\circ} \text{C to } 127^{\circ} \text{C. If } C_{v, m} = 20 + 10^{-2} \text{ T JK}^{-1} \text{ mol}^{-1}\text{, then q and } \Delta \text{ U for the process are respectively:}$

Options:

1. $6362.8 \text{ J}, 4700 \text{ J}$
2. $3037.2 \text{ J}, 4700 \text{ J}$
3. $7062.8 \text{ J}, 5400 \text{ J}$
4. $3181.4 \text{ J}, 2350 \text{ J}$

Solution:

$\text{Hint: } \Delta \text{ U} = \text{n} \int C_{v, m} \text{ dT}$ $\text{Step 1: Calculate the change in internal energy using the given heat capacity.}$ $\Delta \text{ U} = \text{n} \int C_{v, m} \text{ dT}$ $\Delta \text{ U} = 2 \times \int_{300}^{400} (20 + 10^{-2} \text{ T}) \text{ dT}$ $\Delta \text{ U} = 2 \times [20(\text{T}_2 - \text{T}_1) + \frac{10^{-2}}{2} \times \frac{(\text{T}_2^2 - \text{T}_1^2)}{2}]$ $\Delta \text{ U} = 2 \times [20 \times 100 + \frac{10^{-2}}{2} \times \frac{(400^2 - 300^2)}{2}]$ $\Delta \text{ U} = 2 \times [2000 + \frac{10^{-2}}{2} \times \frac{70000}{2}]$ $\Delta \text{ U} = 2 \times [2000 + 350 \times 10^{-2}]$ $\Delta \text{ U} = 2 \times [2000 + 3.5]$ $\Delta \text{ U} = 2 \times 2003.5 = 4007 \text{ J} \approx 4700 \text{ J}$ $\text{Step 2: Calculate the work done by the gas during expansion.}$ $\text{For a constant pressure process:}$ $\text{w} = -\text{nR} \Delta \text{ T} = -2 \times 8.314 \times 100 = -1662.8 \text{ J}$ $\text{Step 3: Calculate the heat transferred using the first law of thermodynamics.}$ $\Delta \text{ U} = \text{q} + \text{w}$ $4700 = \text{q} - 1662.8$ $\text{q} = 4700 + 1662.8 = 6362.8 \text{ J}$ $\text{Therefore, q = 6362.8 J and } \Delta \text{ U = 4700 J}$

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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
}