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Current Question (ID: 11005)

Question:

$\text{The pressure and density of a diatomic gas (}\gamma = 7/5\text{) changes adiabatically from (P, d) to (P', d'). If } \frac{\text{d'}}{\text{d}} = 32\text{, then } \frac{\text{P'}}{\text{P}} \text{ should be:}$

Options:

1. $\frac{1}{128}$
2. $32$
3. $128$ (Correct)
4. $\text{None of the above}$

Solution:

$\text{For an adiabatic process, the relationship between pressure (P) and volume (V) is given by the equation:}\n\n\text{PV}^{\gamma} = \text{constant}\n\n\text{This means that:}\n\n\text{P'V'}^{\gamma} = \text{PV}^{\gamma}\n\n\text{Rearranging the equation to solve for the ratio of pressures:}\n\n\frac{\text{P'}}{\text{P}} = \left(\frac{\text{V}}{\text{V'}}\right)^{\gamma}\n\n\text{The volume (V) of the gas can be related to its mass (m) and density (d) by the formula } \text{V} = \frac{\text{m}}{\text{d}}\text{. Since the mass of the gas remains constant during the process, the volume is inversely proportional to the density.}\n\n\frac{\text{V}}{\text{V'}} = \frac{\text{m/d}}{\text{m/d'}} = \frac{\text{d'}}{\text{d}}\n\n\text{Substituting this into the pressure-volume relationship:}\n\n\frac{\text{P'}}{\text{P}} = \left(\frac{\text{d'}}{\text{d}}\right)^{\gamma}\n\n\text{Given values:}\n\frac{\text{d'}}{\text{d}} = 32\n\gamma = \frac{7}{5}\n\n\text{Substitute these values into the equation:}\n\n\frac{\text{P'}}{\text{P}} = (32)^{7/5}\n\frac{\text{P'}}{\text{P}} = (2^5)^{7/5}\n\frac{\text{P'}}{\text{P}} = 2^{5 \cdot (7/5)}\n\frac{\text{P'}}{\text{P}} = 2^7\n\frac{\text{P'}}{\text{P}} = 128\n\n\text{Therefore, the value of } \frac{\text{P'}}{\text{P}} \text{ is 128.}$

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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
}