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Current Question (ID: 10995)

Question:

$\text{Given below are two statements:}$ $\text{Statement I: When } \mu \text{ amount of an ideal gas undergoes adiabatic change from state } (P_1, V_1, T_1) \text{ to state } (P_2, V_2, T_2)\text{, the work done is } W = \frac{\mu R(T_2 - T_1)}{1 - \gamma}\text{, where } \gamma = \frac{C_P}{C_V} \text{ and } R = \text{universal gas constant.}$ $\text{Statement II: In the above case, when work is done on the gas, the temperature of the gas would rise.}$

Options:

1. $\text{Both Statement I and Statement II are correct.}$ (Correct)
2. $\text{Both Statement I and Statement II are incorrect.}$
3. $\text{Statement I is correct, but Statement II is incorrect.}$
4. $\text{Statement I is incorrect, but Statement II is correct.}$

Solution:

$\text{Hint: } \Delta U = \Delta Q - \Delta W$ $\text{Step: Analyse each statement one by one.}$ $\text{In an adiabatic process, no heat is exchanged } (Q = 0)\text{, so the first law of thermodynamics becomes:}$ $W = -\Delta U = -nC_V(T_2 - T_1)$ $W = -\mu \left(\frac{f}{2}\right) R(T_2 - T_1)$ $\text{We know that: } \frac{2}{f} + 1 = \gamma$ $\frac{f}{2} = \frac{1}{\gamma - 1}$ $W = -\mu \left(\frac{f}{2}\right) R(T_2 - T_1)$ $W = -\mu \left(\frac{1}{\gamma - 1}\right) R(T_2 - T_1) = \frac{\mu R(T_2 - T_1)}{1 - \gamma} \ldots (1)$ $Q = W + \Delta U$ $0 = W + \Delta U$ $\Delta U = -W$ $\text{If work is done on the gas, i.e., work is negative so } \Delta U > 0$ $\text{In an adiabatic process, if work is done on the gas (i.e., the gas is compressed), the internal energy of the gas increases, which raises its temperature. Since there is no heat exchange } (Q = 0)\text{, the increase in internal energy comes entirely from the work done on the gas.}$ $\text{Therefore, both Statement I and Statement II are correct and consistent with the behavior of an ideal gas undergoing an adiabatic process.}$ $\text{Hence, option (1) is the correct answer.}$

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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
}