Import Question JSON

Current Question (ID: 11118)

Question:

$\text{At } 10^{\circ} \text{C the value of the density of a fixed mass of an ideal gas divided by its pressure is } x \text{. At } 110^{\circ} \text{C this ratio is:}$

Options:

1. $x$
2. $\frac{383}{283}x$
3. $\frac{10}{110}x$
4. $\frac{283}{383}x$ (Correct)

Solution:

$\text{We start with the ideal gas equation: } PV = nRT \text{. We can replace the number of moles } n \text{ with } \frac{m}{M} \text{, where } m \text{ is the mass and } M \text{ is the molar mass of the gas. The equation becomes } PV = \frac{m}{M}RT \text{. Rearranging this, we get } \frac{P}{m} = \frac{RT}{VM} \text{ or } \frac{P V}{m} = \frac{RT}{M} \text{. We know that density } \rho = \frac{m}{V} \text{. So, we can rewrite the ideal gas equation in terms of density: } P = \frac{\rho}{M}RT \text{. Rearranging to find the ratio of density to pressure, we get } \frac{\rho}{P} = \frac{M}{RT} \text{. This shows that the ratio of density to pressure is inversely proportional to the absolute temperature, } T \text{, as } M \text{ and } R \text{ are constants for a fixed mass of a gas. So, } \frac{\rho_1}{P_1} = \frac{M}{RT_1} \text{ and } \frac{\rho_2}{P_2} = \frac{M}{RT_2} \text{. Dividing these two equations gives us } \frac{\rho_2/P_2}{\rho_1/P_1} = \frac{T_1}{T_2} \text{. We are given that the initial ratio is } x \text{, so } \frac{\rho_1}{P_1} = x \text{. The initial temperature is } T_1 = 10^{\circ}\text{C} = 10+273 = 283 \text{ K. The final temperature is } T_2 = 110^{\circ}\text{C} = 110+273 = 383 \text{ K. Substituting these values into the ratio equation, we get } \frac{\rho_2/P_2}{x} = \frac{283}{383} \text{. Multiplying both sides by } x \text{, we find the new ratio: } \frac{\rho_2}{P_2} = \frac{283}{383}x \text{. Hence, option (4) is the correct answer.}$

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
}