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Current Question (ID: 11006)

Question:

$\text{At a pressure of 2 atmospheres, a mass of diatomic gas (}\gamma = 1.4\text{), is compressed adiabatically, causing its temperature to rise from } 27^{\circ}\text{C to } 927^{\circ}\text{C. The pressure of the gas in the final state is:}$

Options:

1. $8\text{ atm}$
2. $28\text{ atm}$
3. $68.7\text{ atm}$
4. $256\text{ atm}$ (Correct)

Solution:

$\text{For an adiabatic process, the relationship between pressure (P) and temperature (T) is given by the equation:}\n\n$\text{P} \propto \text{T}^{\gamma/(\gamma-1)}\n\text{This can also be written as:}\n\n\frac{\text{P}_2}{\text{P}_1} = \left(\frac{\text{T}_2}{\text{T}_1}\right)^{\gamma/(\gamma-1)}\n\text{Rearranging the equation to solve for the final pressure (}\text{P}_2\text{):}\n\n\text{P}_2 = \text{P}_1 \left(\frac{\text{T}_2}{\text{T}_1}\right)^{\gamma/(\gamma-1)}\n\n\text{First, convert the temperatures from Celsius to Kelvin:}\n\text{T}_1 = 27^{\circ}\text{C} + 273 = 300\text{ K}\n\text{T}_2 = 927^{\circ}\text{C} + 273 = 1200\text{ K}\n\n\text{Given values:}\n\text{P}_1 = 2\text{ atm}\n\text{T}_1 = 300\text{ K}\n\text{T}_2 = 1200\text{ K}\n\gamma = 1.4\n\n\text{Substitute these values into the equation:}\n\n\text{P}_2 = 2 \left(\frac{1200}{300}\right)^{1.4/(1.4-1)}\n\text{P}_2 = 2 \left(4\right)^{1.4/0.4}\n\text{P}_2 = 2 \left(4\right)^{3.5}\n\text{Since } 4^{3.5} = (4^3) \cdot (4^{0.5}) = 64 \cdot 2 = 128\n\text{P}_2 = 2 \cdot 128 = 256\text{ atm}\n\n\text{Therefore, the final pressure of the gas is 256 atm.}$

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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
}