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Current Question (ID: 8288)

Question:

$\text{A piston filled with 0.04 mol of an ideal gas expands reversibly from 50.0 mL to 375 mL at a constant temperature of } 37.0^{\circ}\text{C. As it does so, it absorbs 208 J of heat. The values of q and w for the process will be-}$ $(\text{R} = 8.314 \text{ J/mol K}) (\ln 7.5 = 2.01)$

Options:

1. $\text{q} = +208 \text{ J}, \text{w} = -208 \text{ J}$
2. $\text{q} = -208 \text{ J}, \text{w} = -208 \text{ J}$
3. $\text{q} = -208 \text{ J}, \text{w} = + 208 \text{ J}$
4. $\text{q} = +208 \text{ J}, \text{w} = + 208 \text{ J}$

Solution:

$\text{Hint: Use the First law of thermodynamics}$ $\text{Step 1: Analyze the process}$ $\text{The process is an isothermal expansion. For an isothermal process, the temperature remains constant:}$ $\Delta \text{U} = 0$ $\text{We can verify this using the relation: } \Delta \text{U} = C_V \Delta \text{T}$ $\text{Since } \Delta \text{T} = 0 \text{ (isothermal process), } \Delta \text{U} = 0$ $\text{Step 2: Apply the First Law of Thermodynamics}$ $\text{From the First Law of Thermodynamics: } \Delta \text{U} = \text{q} + \text{w}$ $\text{Since } \Delta \text{U} = 0\text{:}$ $0 = \text{q} + \text{w}$ $\text{Therefore: } \text{q} = -\text{w}$ $\text{Step 3: Determine the values of q and w}$ $\text{The problem states that the gas absorbs 208 J of heat during the expansion.}$ $\text{When a system absorbs heat, q is positive according to the sign convention:}$ $\text{q} = +208 \text{ J}$ $\text{Using } \text{q} = -\text{w}\text{:}$ $\text{w} = -\text{q} = -(+208 \text{ J}) = -208 \text{ J}$ $\text{The negative sign for work indicates that the system is doing work on the surroundings (expansion work).}$ $\text{We can verify this result using the formula for isothermal reversible expansion work:}$ $\text{w} = -\text{nRT} \ln\frac{V_2}{V_1} = -0.04 \times 8.314 \times 310 \times \ln\frac{375}{50} = -0.04 \times 8.314 \times 310 \times \ln(7.5)$ $\text{w} = -0.04 \times 8.314 \times 310 \times 2.01 \approx -208 \text{ J}$ $\text{Therefore, q = +208 J and w = -208 J, corresponding to option 1.}$

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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
}