Import Question JSON

Current Question (ID: 11125)

Question:

$\text{Two closed containers of equal volume are filled with air at pressure } P_0 \text{ and temperature } T_0. \text{ Both are connected by a narrow tube. If one of the containers is maintained at temperature } T_0 \text{ and the other at temperature } T, \text{ then new pressure in the container will be:}$

Options:

1. $\frac{2P_0 T}{T+T_0}$ (Correct)
2. $\frac{P_0 T}{T+T_0}$
3. $\frac{P_0 T}{2(T+T_0)}$
4. $\frac{T+T_0}{P_0}$

Solution:

$\text{Hint: The number of moles is conserved. Step 1: Find the initial number of moles. Using the ideal gas equation } PV=nRT, \text{ the initial number of moles in each container is } n = \frac{P_0 V_0}{RT_0}. \text{ The total initial number of moles is } n_i = n + n = 2n = \frac{2P_0 V_0}{RT_0}. \text{ Step 2: Find the final number of moles. The final pressure } P' \text{ will be the same in both containers because they are connected. The volumes are } V_0 \text{ for each. The final number of moles } n_f \text{ is the sum of moles in each container: } n_f = n_1 + n_2 = \frac{P' V_0}{RT} + \frac{P' V_0}{RT_0} = \frac{P'V_0}{R} \left( \frac{1}{T} + \frac{1}{T_0} \right) = \frac{P'V_0}{R} \left( \frac{T_0 + T}{TT_0} \right). \text{ Step 3: Conserve the number of moles. Since the number of moles is conserved, we set } n_i = n_f. \text{ So, } \frac{2P_0 V_0}{RT_0} = \frac{P'V_0}{R} \left( \frac{T_0 + T}{TT_0} \right). \text{ Cancel out } \frac{V_0}{R} \text{ from both sides: } \frac{2P_0}{T_0} = P' \left( \frac{T_0 + T}{TT_0} \right). \text{ Rearrange to solve for } P': P' = \frac{2P_0}{T_0} \left( \frac{TT_0}{T+T_0} \right) = \frac{2P_0 T}{T+T_0}.\text{ Thus, option 1 is correct.}$

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
}