Import Question JSON

Current Question (ID: 8123)

Question:

$\text{A plot of volume versus temperature (T) for a gas at constant pressure is a straight line passing through the origin. The plots at different values of pressure are shown in the figure given below.}$ $\text{The correct order of pressure is -}$

Options:

1. $p_1 > p_2 > p_3 > p_4$
2. $p_1 = p_2 = p_3 = p_4$
3. $p_1 < p_2 < p_3 < p_4$
4. $p_1 < p_2 = p_3 < p_4$

Solution:

$\text{According to Boyle's law, at a particular temperature,}$ $PV = \text{constant}$ $\text{So,}$ $P = \text{constant} / V$ $\text{Also, from the ideal gas equation,}$ $PV = nRT$ $\text{From the graph, it is clear that for a given volume, say } V_1 \text{, the corresponding temperatures are } T_1, T_2, T_3, T_4 \text{ for pressures } P_1, P_2, P_3, P_4 \text{ respectively. }$ $\text{We can observe that } T_1 < T_2 < T_3 < T_4 \text{.}$ $\text{Since } P = \text{constant} / V \text{ and } T \text{ is directly proportional to } V \text{ (from Charles' Law } V \propto T \text{ at constant P),}$ $\text{Thus, for a given volume, as temperature increases, pressure decreases.}$ $\text{So, for a fixed volume, since } T_1 < T_2 < T_3 < T_4 \text{, it implies } P_1 > P_2 > P_3 > P_4 \text{.}$ $\text{Alternatively, consider a constant temperature (vertical line). As we move from } P_1 \text{ to } P_4 \text{, the volume increases.}$ $\text{Since } PV = \text{constant (at constant T), if V increases, P must decrease.}$ $\text{Therefore, } P_1 > P_2 > P_3 > P_4 \text{.}$

Import JSON File

Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}