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Current Question (ID: 19675)

Question:

$\text{Match the thermodynamic processes taking place in a system with the correct conditions. In the table: } \Delta Q \text{ is the heat supplied, } \Delta W \text{ is the work done and } \Delta U \text{ is change in internal energy of the system.}$ $\begin{array}{|c|c|} \hline \text{Process} & \text{Condition} \\ \hline (\text{I}) \text{ Adiabatic} & (\text{A}) \Delta W = 0 \\ (\text{II}) \text{ Isothermal} & (\text{B}) \Delta Q = 0 \\ (\text{III}) \text{ Isochoric} & (\text{C}) \Delta U \neq 0, \Delta W \neq 0, \Delta Q \neq 0 \\ (\text{IV}) \text{ Isobaric} & (\text{D}) \Delta U = 0 \\ \hline \end{array}$

Options:

1. $(\text{I}) - (\text{B}), (\text{II}) - (\text{A}), (\text{III}) - (\text{D}), (\text{IV}) - (\text{C})$
2. $(\text{I}) - (\text{A}), (\text{II}) - (\text{B}), (\text{III}) - (\text{C}), (\text{IV}) - (\text{D})$
3. $(\text{I}) - (\text{A}), (\text{II}) - (\text{B}), (\text{III}) - (\text{D}), (\text{IV}) - (\text{D})$
4. $(\text{I}) - (\text{B}), (\text{II}) - (\text{D}), (\text{III}) - (\text{A}), (\text{IV}) - (\text{C})$

Solution:

$\text{Hint: In adiabatic process, } \Delta Q = 0.$ $\text{(I) Adiabatic process } \Rightarrow \Delta Q = 0$ $\text{No exchange of heat takes place with surroundings}$ $\text{(II) Isothermal process } \Rightarrow \text{Temperature remains constant } (\Delta T = 0)$ $\Delta u = \frac{F}{2} n R \Delta T \Rightarrow \Delta U = 0$ $\text{No change in internal energy } [\Delta u = 0]$ $\text{(III) Isochoric process Volume remains constant}$ $\Delta V = 0$ $W = \int P \cdot dV = 0$ $\text{Hence work done is zero.}$ $\text{(IV) Isobaric process } \Rightarrow \text{Pressure remains constant}$ $W = P \cdot \Delta V \neq 0$ $\Delta u = \frac{F}{2} n R \Delta T = \frac{F}{2} [P \Delta V] \neq 0$ $\Delta Q = n C_p \Delta T \neq 0$

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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
}