Import Question JSON

Current Question (ID: 11130)

Question:

$\text{At constant temperature, on increasing the pressure of a gas by 5\%, its volume will decrease by:}$

Options:

1. $\text{5\%}$
2. $\text{5.26 \%}$
3. $\text{4.26 \%}$
4. $\text{4.76 \%}$ (Correct)

Solution:

$\text{According to Boyle's law, at constant temperature, the pressure of a given mass of an ideal gas is inversely proportional to its volume.}\text{This can be represented as: }\text{P} \propto \frac{1}{\text{V}} \text{ or } \text{P}_1\text{V}_1 = \text{P}_2\text{V}_2.\text{Let the initial pressure be }\text{P}_1\text{ and the initial volume be }\text{V}_1.\text{The pressure is increased by 5\%, so the new pressure }\text{P}_2\text{ is: }\text{P}_2 = \text{P}_1 + 0.05\text{P}_1 = 1.05\text{P}_1.\text{Using Boyle's Law: }\text{P}_1\text{V}_1 = \text{P}_2\text{V}_2 \Rightarrow \text{P}_1\text{V}_1 = (1.05\text{P}_1)\text{V}_2.\text{Solving for the new volume }\text{V}_2\text{: }\text{V}_2 = \frac{\text{P}_1\text{V}_1}{1.05\text{P}_1} = \frac{1}{1.05}\text{V}_1 \approx 0.95238\text{V}_1.\text{The percentage decrease in volume is given by the formula: }\text{Percentage Decrease} = \frac{\text{Initial Volume} - \text{Final Volume}}{\text{Initial Volume}} \times 100.\text{Percentage Decrease} = \frac{\text{V}_1 - \text{V}_2}{\text{V}_1} \times 100 = \frac{\text{V}_1 - 0.95238\text{V}_1}{\text{V}_1} \times 100 = (1 - 0.95238) \times 100 = 0.04762 \times 100 \approx 4.76\%.\text{Therefore, the volume decreases by approximately 4.76\%.}$

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
}