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Current Question (ID: 18951)

Question:

$\text{Pressure } (P), \text{ volume } (V), \text{ temperature } (T) \text{ are related to each other as per the following relation:}$ $\left( P + \frac{a}{V^2} \right) (V-b) = RT$ $\text{Dimensions of } \frac{a}{b} \text{ are:}$

Options:

1. $[MLT^{-2}]$
2. $[M^2LT^{-2}]$
3. $[ML^2T^{-2}]$
4. $[ML^3T^{-1}]$

Solution:

$\text{Hint: Use the principle of dimensional homogeneity.}$ $\text{Step 1: Find the dimensions of } a \text{ and } b.$ $\text{As, } \frac{a}{V^2} = \text{Correction in pressure}$ $\left[ \frac{a}{V^2} \right] = [\text{Pressure}] = [ML^{-1}T^{-2}]$ $\Rightarrow [a] = [ML^{-1}T^{-2}][L^6] = [ML^5T^{-2}]$ $\text{Similarly, } b = \text{Correction in volume}$ $\text{Then, } [b] = [\text{Volume}] = [L^3]$ $\text{Step 2: Find the dimension of } \frac{a}{b}.$ $\left[ \frac{a}{b} \right] = \frac{[ML^5T^{-2}]}{[L^3]} = [ML^2T^{-2}]$ $\text{Hence, option (3) is the correct answer.}$

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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
}