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Current Question (ID: 19838)

Question:

$\text{One mole of an ideal gas undergoes a process in which pressure and volume are related by the equation:}$ $P = P_0 \left[ 1 - \frac{1}{2} \left( \frac{V_0}{V} \right)^2 \right]$ $\text{where } P_0 \text{ and } V_0 \text{ are constants. If the volume of the gas increases from } V = V_0 \text{ to } V = 2V_0, \text{ what is the resulting change in temperature?}$

Options:

1. $\frac{3}{4} \frac{P_0 V_0}{R}$
2. $\frac{1}{2} \frac{R}{P_0 V_0}$
3. $\frac{5}{4} \frac{R}{P_0 V_0}$
4. $\frac{1}{4} \frac{P_0 V_0}{R}$

Solution:

$\text{Hint: } PV = nRT$ $P = P_0 \left[ 1 - 2 \left( \frac{V_0}{V} \right)^2 \right]$ $\text{When } V_1 = V_0$ $V_2 = 2V_0$ $\Delta T = T_2 - T_1 = (P_2 V_2 - V_1 P_1) \frac{1}{nR}$ $= \left( \frac{7}{4} - \frac{1}{2} \right) \frac{P_0 V_0}{R}$ $= \frac{5}{4} \frac{P_0 V_0}{R}$

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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
}