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Current Question (ID: 11081)

Question:

$\text{In the } (P-V) \text{ diagram shown, the gas does 5 J of work in the isothermal process } ab \text{ and 4 J in the adiabatic process } bc\text{. What will be the change in internal energy of the gas in the straight path from } c \text{ to } a\text{?}$

Options:

1. $9 \text{ J}$
2. $1 \text{ J}$
3. $4 \text{ J}$ (Correct)
4. $5 \text{ J}$

Solution:

$\text{Hint: The isothermal process is the process in which the temperature of the system remains constant.}$ $\text{Step: Find the change in internal energy of the gas in the straight path from } c \text{ to } a$ $\text{From the diagram, we can understand that the process is cyclic. Therefore the net change in internal energy between the processes } a \text{ to } b \text{ to } c \text{ and } c \text{ equals to zero.}$ $\Delta U = \Delta U_a + \Delta U_b + \Delta U_c = 0 \quad \ldots (1)$ $\text{For isothermal: } \Delta U_a = 0$ $\text{For adiabatic: } \Delta U = -\Delta W$ $\text{As we know the work is done in the process } b - c \text{ is 4 J}$ $\rightarrow \Delta U_b = -4 \text{ J}$ $\text{Applying this value in the equation (1), we get}$ $\Delta U_c = 4 \text{ J}$ $\text{Thus the change in internal energy of the gas in the straight path } c \text{ to } a \text{ is 4 J.}$ $\text{Hence, option (3) is the correct answer.}$

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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
}