Import Question JSON

Current Question (ID: 19681)

Question:

$\text{If one mole of an ideal gas at } (p_1, V_1) \text{ is allowed to expand reversibly and isothermally } (A \rightarrow B), \text{ its pressure is reduced to one-half of the original pressure (see figure).}$ $\text{This is followed by a constant-volume cooling till its pressure is reduced to one-fourth of the initial value } (B \rightarrow C). \text{ Then it is restored to its initial state by a reversible adiabatic compression}(C \rightarrow A). \text{ The net work done by the gas is:}$

Options:

1. $RT \left( \ln 2 - \frac{1}{2(\gamma - 1)} \right)$
2. $-\frac{RT}{2(\gamma - 1)}$
3. $0$
4. $RT \ln 2$

Solution:

$\text{Hint: Work done in an isochoric process is zero.}$ $\text{Step: Find the net work done by the gas.}$ $W_{\text{Net}} = W_{AB} + W_{BC} + W_{CA}$ $\Rightarrow W_{\text{Net}} = nRT \ln \left( \frac{V_f}{V_i} \right) + 0 + \frac{P_i V_i - P_f V_f}{\gamma - 1}$ $\Rightarrow W_{\text{Net}} = nRT \ln \left( \frac{2V}{V} \right) + 0 + \frac{\frac{P_1}{4} \times 2V_1 - P_1 V_1}{\gamma - 1}$ $\Rightarrow W_{\text{Net}} = nRT \ln \left( \frac{2V}{V} \right) + 0 - \frac{P_1 V_1}{2(\gamma - 1)}$ $\Rightarrow W_{\text{Net}} = nRT \ln(2) + 0 - \frac{nRT}{2(\gamma - 1)}$ $\Rightarrow W_{\text{Net}} = nRT \left( \ln(2) - \frac{1}{2(\gamma - 1)} \right)$ $\text{Hence, option (1) is the correct answer.}$

Import JSON File

Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}