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Current Question (ID: 8230)

Question:

$\text{Choose the correct statement for viscosity } (\eta) \text{ variation with T and P for an ideal gas-}$

Options:

1. $\eta \text{ of a gas increases with increase in temperature (T), but it is independent of pressure}$
2. $\eta \text{ of a gas decreases with increase in T and increases with increase in P}$
3. $\eta \text{ of a gas is independent of temperature and pressure}$
4. $\eta \text{ of a gas increases with increase in both T and P}$

Solution:

$\text{Hint: } F = \eta A \frac{du}{dz}$ $\text{In the case of gases, the viscous resistance arises from the transport of molecules from one layer to another, with transfer of momenta, so that the fast moving molecules in one layer are slowed down, while the slow moving molecules in the other are accelerated. This momentum transfer increases with increase in temperature. A simple expression for the viscosity of a gas}$ $\eta = \frac{1}{3} \overline{C} \text{ dl, where d is the density and l and the mean free path.}$ $\text{d} \propto \text{pressure and l} \propto \frac{1}{\text{pressure}} \text{ and so } \eta \text{ is independent of pressure.}$ $\text{Therefore, viscosity of a gas increases with temperature but is independent of pressure.}$

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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
}