Import Question JSON

Current Question (ID: 11131)

Question:

$\text{A thermally insulated piston divides a container into two compartments. Volume, temperature, and pressure in the right compartment are } 2\text{V, T and } 2\text{P, while in the left compartment the respective values are V, T and P. If the piston can slide freely, then in the final equilibrium position, the volume of the right-hand compartment will be:}$

Options:

1. $\text{V}$
2. $\frac{3\text{V}}{5}$
3. $\frac{9\text{V}}{4}$
4. $\frac{12\text{V}}{5}$ (Correct)

Solution:

$\text{We are given the initial conditions for both compartments.}\text{Let the number of moles in the left compartment be }\text{n}_1\text{ and in the right compartment be }\text{n}_2\text{.}\text{From the ideal gas equation, }\text{PV} = \text{nRT}.\text{For the left compartment: }\text{PV} = \text{n}_1\text{RT} \Rightarrow \text{n}_1 = \frac{\text{PV}}{\text{RT}}.\text{For the right compartment: }\text{P}_2\text{V}_2 = \text{n}_2\text{RT}_2 \Rightarrow \text{(2P)(2V)} = \text{n}_2\text{RT} \Rightarrow 4\text{PV} = \text{n}_2\text{RT} \Rightarrow \text{n}_2 = \frac{4\text{PV}}{\text{RT}}.\text{This shows that }\text{n}_2 = 4\text{n}_1\text{, or the right compartment has 4 times the number of moles as the left compartment.}\text{When the piston slides freely, the final pressure on both sides will be equal. Let the final pressure be }\text{P}'\text{.}\text{The total volume of the container is constant: }\text{V}_{\text{total}} = \text{V}_{\text{left}} + \text{V}_{\text{right}} = \text{V} + 2\text{V} = 3\text{V}.\text{In the final state, let the new volumes be }\text{V}'_{L}\text{ and }\text{V}'_{R}\text{ for the left and right compartments, respectively.}\text{So, }\text{V}'_{L} + \text{V}'_{R} = 3\text{V}.\text{The temperature remains constant since the piston is thermally insulated.}\text{For the left compartment: }\text{P}'\text{V}'_{L} = \text{n}_1\text{RT} = \text{PV} \Rightarrow \text{V}'_{L} = \frac{\text{PV}}{\text{P}'}.\text{For the right compartment: }\text{P}'\text{V}'_{R} = \text{n}_2\text{RT} = 4\text{PV} \Rightarrow \text{V}'_{R} = \frac{4\text{PV}}{\text{P}'}.\text{Substituting these into the total volume equation: }\frac{\text{PV}}{\text{P}'} + \frac{4\text{PV}}{\text{P}'} = 3\text{V}.\text{Multiplying by }\frac{\text{P}'}{\text{V}}\text{ gives: }\text{P} + 4\text{P} = 3\text{P}' \Rightarrow 5\text{P} = 3\text{P}' \Rightarrow \text{P}' = \frac{5\text{P}}{3}.\text{We need to find the final volume of the right compartment, }\text{V}'_{R}\text{.}\text{Using the equation for the right compartment: }\text{V}'_{R} = \frac{4\text{PV}}{\text{P}'} = \frac{4\text{PV}}{\frac{5\text{P}}{3}} = \frac{4\text{PV} \cdot 3}{5\text{P}} = \frac{12\text{V}}{5}.\text{Thus, the volume of the right-hand compartment will be }\frac{12\text{V}}{5}.\text{The correct option is 4.}$

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
}