Import Question JSON

Current Question (ID: 11117)

Question:

$\text{Two thermally insulated vessels } 1 \text{ and } 2 \text{ are filled with air at temperatures } T_1, T_2 \text{, volume } V_1, V_2 \text{ and pressure } P_1, P_2 \text{ respectively. If the valve joining the two vessels is opened, the temperature inside the vessel at equilibrium will be:}$

Options:

1. $T_1 + T_2$
2. $\frac{T_1 + T_2}{2}$
3. $\frac{T_1 T_2 (P_1 V_1 + P_2 V_2)}{P_1 V_1 T_2 + P_2 V_2 T_1}$ (Correct)
4. $\frac{T_1 T_2 (P_1 V_1 + P_2 V_2)}{P_1 V_1 T_1 + P_2 V_2 T_2}$

Solution:

$\text{Since the vessels are thermally insulated, the total number of moles of the gas remains constant. According to the ideal gas law, } PV = nRT \text{, which can be rearranged to find the number of moles } n = \frac{PV}{RT} \text{. The total number of moles before opening the valve is the sum of the moles in each vessel: } n_\text{total} = n_1 + n_2 = \frac{P_1 V_1}{RT_1} + \frac{P_2 V_2}{RT_2} \text{. When the valve is opened and the system reaches equilibrium, the total volume is } V = V_1 + V_2 \text{ and the final pressure is } P \text{. The final temperature is } T \text{. The total number of moles at equilibrium is } n_\text{total} = \frac{P(V_1+V_2)}{RT} \text{. Since the total number of moles is conserved, we can set the initial and final moles equal to each other: } \frac{P_1 V_1}{RT_1} + \frac{P_2 V_2}{RT_2} = \frac{P(V_1+V_2)}{RT} \text{. The solution provided seems to have an error by applying Boyle's law which is for a fixed temperature system. The correct approach is based on the conservation of the number of moles and the ideal gas law. From the ideal gas law, we have } n_1 = \frac{P_1 V_1}{RT_1} \text{ and } n_2 = \frac{P_2 V_2}{RT_2} \text{. After opening the valve, the new number of moles is } n = \frac{P(V_1+V_2)}{RT} \text{, where } T \text{ is the final temperature. Since } n = n_1 + n_2 \text{, we have } \frac{P(V_1+V_2)}{RT} = \frac{P_1 V_1}{RT_1} + \frac{P_2 V_2}{RT_2} \text{. The common denominator on the right side is } RT_1 T_2 \text{, so } \frac{P(V_1+V_2)}{RT} = \frac{P_1 V_1 T_2 + P_2 V_2 T_1}{RT_1 T_2} \text{. By canceling out } R \text{, we have } \frac{P(V_1+V_2)}{T} = \frac{P_1 V_1 T_2 + P_2 V_2 T_1}{T_1 T_2} \text{. Now, let's solve for } T \text{. } T = \frac{P(V_1+V_2)T_1 T_2}{P_1 V_1 T_2 + P_2 V_2 T_1} \text{. However, we do not know the final pressure } P \text{. This suggests an issue with the premise of the solution provided. The solution provided leads to option 3, which is a known formula for this type of problem. Let's re-examine the derivation. The solution seems to have made a simplification by incorrectly assuming Boyle's law applies. Instead, let's assume the question implicitly asks for the temperature without needing the final pressure. The formula provided in option 3 is } \frac{T_1 T_2 (P_1 V_1 + P_2 V_2)}{P_1 V_1 T_2 + P_2 V_2 T_1} \text{. Let's check if this can be derived. The total internal energy is conserved, and for an ideal gas, internal energy is a function of temperature. The change in internal energy } \Delta U = nC_V \Delta T \text{ is zero, so } n_1 C_V T_1 + n_2 C_V T_2 = (n_1+n_2)C_V T \text{ which simplifies to } n_1 T_1 + n_2 T_2 = (n_1+n_2)T \text{. Thus, } T = \frac{n_1 T_1 + n_2 T_2}{n_1+n_2} \text{. Substituting } n_1 = \frac{P_1 V_1}{RT_1} \text{ and } n_2 = \frac{P_2 V_2}{RT_2} \text{, we get } T = \frac{(\frac{P_1 V_1}{RT_1})T_1 + (\frac{P_2 V_2}{RT_2})T_2}{(\frac{P_1 V_1}{RT_1}) + (\frac{P_2 V_2}{RT_2})} = \frac{\frac{P_1 V_1}{R} + \frac{P_2 V_2}{R}}{\frac{P_1 V_1 T_2 + P_2 V_2 T_1}{RT_1 T_2}} \text{. Simplifying this, we get } T = \frac{P_1 V_1 + P_2 V_2}{R} \cdot \frac{RT_1 T_2}{P_1 V_1 T_2 + P_2 V_2 T_1} = \frac{(P_1 V_1 + P_2 V_2) T_1 T_2}{P_1 V_1 T_2 + P_2 V_2 T_1} \text{. This matches option 3.}$

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
}