Import Question JSON

Current Question (ID: 11008)

Question:

$\text{The volume of air (diatomic) increases by } 5\% \text{ in its adiabatical expansion. The percentage decrease in its pressure will be:}$

Options:

1. $5\%$
2. $6\%$
3. $7\%$ (Correct)
4. $8\%$

Solution:

$\text{For an adiabatic process, the relationship between pressure } (P) \text{ and volume } (V) \text{ is given by:}$ \n\n $PV^{\gamma} = K \text{ (where } K \text{ is a constant)}$ \n\n $\text{Taking the logarithm of both sides:}$ \n\n $\text{ln}(P) + \gamma \text{ln}(V) = \text{ln}(K)$ \n\n $\text{Differentiating the equation with respect to } P \text{ and } V \text{ we get:}$ \n\n $\frac{dP}{P} + \gamma \frac{dV}{V} = 0$ \n\n $\text{This can be rewritten as:}$ \n\n $\frac{dP}{P} = -\gamma \frac{dV}{V}$ \n\n $\text{To find the percentage change, we multiply by } 100\%\text{:}$ \n\n $\frac{dP}{P} \times 100 = -\gamma (\frac{dV}{V} \times 100)$ \n\n $\text{Air is considered a diatomic gas. For a diatomic gas, the adiabatic index } (\gamma) \text{ is approximately } 1.4\text{.}$ \n\n $\text{The percentage increase in volume is given as } 5\%\text{. So, } (\frac{dV}{V} \times 100) = 5\%\text{.}$ \n\n $\text{Substituting these values into the equation:}$ \n\n $\frac{dP}{P} \times 100 = -1.4 \times 5\%$ \n\n $\frac{dP}{P} \times 100 = -7\%$ \n\n $\text{The negative sign indicates a decrease in pressure. Therefore, the percentage decrease in pressure is } 7\%\text{.}$ \n\n $\text{The correct option is 3.}$

Import JSON File

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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
}