Import Question JSON

Current Question (ID: 10999)

Question:

$\text{The heat taken by a gas (shown in the graph below) in the process } a \text{ to } b \text{ is } 6P_0V_0\text{. Match Column I with Column II.}$

Options:

1. $\text{A-P, B-Q, C-S, D-R}$
2. $\text{A-R, B-Q, C-S, D-P}$
3. $\text{A-Q, B-R, C-P, D-S}$
4. $\text{A-Q, B-R, C-S, D-P}$ (Correct)

Solution:

$\text{Hint: Apply the first law of thermodynamics.}$ $\text{Step: Find the correct match.}$ $\text{The work done in the process } ab \text{ i.e., } W_{ab} \text{ is given by:}$ $W_{ab} = \frac{1}{2}[(2P_0 + P_0)(2V_0 - V_0)]$ $\Rightarrow W_{ab} = \frac{3}{2}P_0V_0$ $\text{The internal energy during the process } ab \text{ i.e., } \Delta U_{ab} \text{ is given by:}$ $\Delta U_{ab} = Q_{ab} - W_{ab} = 6P_0V_0 - 1.5P_0V_0$ $\Rightarrow \Delta U_{ab} = \frac{9}{2}P_0V_0 = 4.5P_0V_0$ $\text{The molar heat capacity in the given process is given by:}$ $Q = nC\Delta T$ $\Rightarrow C = \frac{Q}{n\Delta T} = \frac{Q}{3P_0V_0/R}$ $\Rightarrow C = 2R$ $\text{The specific heat at constant volume } C_V \text{ for the gas is given by:}$ $\Delta U = nC_V\Delta T$ $\Rightarrow C_V = \frac{\Delta U}{n\Delta T} = \frac{4.5P_0V_0}{3P_0V_0/R}$ $\Rightarrow C_V = 1.5R$ $\text{Therefore, the correct match is A-Q, B-R, C-S, D-P.}$ $\text{Hence, option (4) is the correct answer.}$

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
}