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Current Question (ID: 11129)

Question:
$\text{Two chambers of different volumes, one containing }\text{m}_1\text{g of a gas at pressure }\text{P}_1\text{ and other containing }\text{m}_2\text{g of the same gas at pressure }\text{P}_2\text{ are joined to each other. If the temperature of the gas remains constant, the common pressure reached is:}$
Options:
  • 1. $\frac{\text{m}_1\text{P}_1 + \text{m}_2\text{P}_2}{\text{m}_1 + \text{m}_2}$
  • 2. $\frac{\text{m}_1\text{P}_2 + \text{m}_2\text{P}_1}{\text{m}_1 + \text{m}_2}$
  • 3. $\frac{\text{m}_1\text{P}_2 + (\text{P}_1+\text{P}_2)}{\text{m}^{2}_1 + \text{m}_2}$
  • 4. $\frac{(\text{m}_1+\text{m}_2)\text{P}_1\text{P}_2}{\text{m}_1\text{P}_2 + \text{m}_2\text{P}_1}$ (Correct)
Solution:
$\text{Let the initial volumes of the two chambers be }\text{V}_1\text{ and }\text{V}_2\text{. The temperature }\text{T}\text{ is constant.}\text{Let the molar mass of the gas be }\text{M}.\text{The number of moles in the first chamber is }\text{n}_1 = \frac{\text{m}_1}{\text{M}}\text{.}\text{The number of moles in the second chamber is }\text{n}_2 = \frac{\text{m}_2}{\text{M}}\text{.}\text{Using the ideal gas equation }\text{PV} = \text{nRT}:\text{For the first chamber: }\text{P}_1\text{V}_1 = \text{n}_1\text{RT} = \frac{\text{m}_1}{\text{M}}\text{RT} \Rightarrow \text{V}_1 = \frac{\text{m}_1\text{RT}}{\text{MP}_1}.\text{For the second chamber: }\text{P}_2\text{V}_2 = \text{n}_2\text{RT} = \frac{\text{m}_2}{\text{M}}\text{RT} \Rightarrow \text{V}_2 = \frac{\text{m}_2\text{RT}}{\text{MP}_2}.\text{When the two chambers are joined, the total volume becomes }\text{V}_{\text{total}} = \text{V}_1 + \text{V}_2\text{ and the total number of moles becomes }\text{n}_{\text{total}} = \text{n}_1 + \text{n}_2\text{.}\text{The common final pressure is }\text{P}.\text{Applying the ideal gas equation to the combined system: }\text{P}(\text{V}_1 + \text{V}_2) = (\text{n}_1 + \text{n}_2)\text{RT}.\text{Substituting the expressions for }\text{V}_1, \text{V}_2, \text{n}_1, \text{ and }\text{n}_2\text{:}\text{P}\left(\frac{\text{m}_1\text{RT}}{\text{MP}_1} + \frac{\text{m}_2\text{RT}}{\text{MP}_2}\right) = \left(\frac{\text{m}_1}{\text{M}} + \frac{\text{m}_2}{\text{M}}\right)\text{RT}.\text{Since RT/M is a common factor on both sides, we can cancel it out:}\text{P}\left(\frac{\text{m}_1}{\text{P}_1} + \frac{\text{m}_2}{\text{P}_2}\right) = \text{m}_1 + \text{m}_2.\text{To solve for }\text{P}, \text{ first combine the fractions on the left side:}\text{P}\left(\frac{\text{m}_1\text{P}_2 + \text{m}_2\text{P}_1}{\text{P}_1\text{P}_2}\right) = \text{m}_1 + \text{m}_2.\text{Now, isolate }\text{P}\text{ by multiplying both sides by the reciprocal of the fraction:}\text{P} = \frac{(\text{m}_1 + \text{m}_2)\text{P}_1\text{P}_2}{\text{m}_1\text{P}_2 + \text{m}_2\text{P}_1}.\text{This matches option 4.}$

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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
}