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Current Question (ID: 11119)

Question:

$\text{The volume and temperature graph is given in the figure below. If pressures for the two processes are different, then which one, of the following, is true?}$

Options:

1. $P_1 = P_2 \text{ and } P_3 = P_4 \text{ and } P_3 > P_2$
2. $P_1 = P_2 \text{ and } P_3 = P_4 \text{ and } P_3 < P_2$ (Correct)
3. $P_1 = P_2 = P_3 = P_4$
4. $P_1 > P_2 > P_3 > P_4$

Solution:

$\text{The graph between } V \text{ and } T \text{ at constant pressure is known as an isobar, and it always gives a straight line. From the provided graph, we can see that the points } P_1 \text{ and } P_2 \text{ lie on the same straight line originating from the origin, which indicates they have the same pressure. Therefore, } P_1 = P_2 \text{. Similarly, the points } P_3 \text{ and } P_4 \text{ lie on another straight line originating from the origin, which indicates they also have the same pressure. Therefore, } P_3 = P_4 \text{. According to the ideal gas law, } PV = nRT \text{, which can be rearranged to } V = \frac{nR}{P}T \text{. This is the equation of a straight line, } y=mx \text{, where the slope is } m = \frac{nR}{P} \text{. From the graph, we can observe that the line for } P_3 \text{ and } P_4 \text{ is steeper than the line for } P_1 \text{ and } P_2 \text{. A steeper slope means a larger value for } \frac{nR}{P} \text{. Since } nR \text{ is constant for a given amount of gas, a larger slope corresponds to a smaller pressure. Therefore, } \frac{nR}{P_3} > \frac{nR}{P_2} \text{, which implies } \frac{1}{P_3} > \frac{1}{P_2} \text{, and thus } P_2 > P_3 \text{. Combining all these observations, the correct statement is that } P_1 = P_2 \text{ and } P_3 = P_4 \text{ and } P_3 < P_2 \text{, which is the same as } P_2 > P_3 \text{.}$

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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
}