Import Question JSON

Current Question (ID: 19842)

Question:

$\text{To raise the temperature of a certain mass of gas by } 50^\circ \text{C at a constant pressure, } 160 \text{ calories of heat is required.}$ $\text{When the same mass of gas is cooled by } 100^\circ \text{C at constant volume, } 240 \text{ calories of heat is released.}$ $\text{How many degrees of freedom does each molecule of this gas have (assume the gas to be ideal)?}$

Options:

1. $2$
2. $5$
3. $6$
4. $3$

Solution:

$\text{Hint: } f = \frac{2}{\gamma - 1}$ $\text{Step: Find the degrees of freedom does each molecule of this gas.}$ $\text{We are given the heat required to raise the temperature by } 50^\circ \text{C at constant pressure is;} \Rightarrow Q_P = 160 \text{ cal}$ $\text{Using } Q = nC_P \Delta T, \text{ we get;} \Rightarrow nC_P(50) = 160$ $\text{The heat released when cooled by } 100^\circ \text{C at constant volume is;} \Rightarrow Q_V = 240 \text{ cal}$ $\text{Using } Q = nC_V \Delta T, \text{ we get;} \Rightarrow nC_V(100) = 240$ $\text{The ratio of specific heats is given by;} \Rightarrow \gamma = \frac{C_P}{C_V} \Rightarrow \frac{3.2}{2.4} = \frac{4}{3}$ $\text{The degrees of freedom does each molecule of this gas is given by;} \Rightarrow f = \frac{2}{\gamma - 1}$ $\text{Substitute the given values we get;} \Rightarrow f = \frac{2}{(4/3) - 1} = 6$ $\text{Hence, option (3) is the correct answer.}$

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
}