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Current Question (ID: 11011)

Question:
$\text{The figure shows the } (P-V) \text{ diagram of an ideal gas undergoing a change of state from } A \text{ to } B\text{. Four different paths I, II, III and IV, as shown in the figure, may lead to the same change of state.}$ $\text{Which of the following options contains only correct statements?}$ $\text{(a) The change in internal energy is the same in cases IV and III but not in cases I and II.}$ $\text{(b) The change in internal energy is the same in all four cases.}$ $\text{(c) The work done is maximum in case I.}$ $\text{(d) The work done is minimum in case II.}$
Options:
  • 1. $\text{(b), (c) and (d) only}$
  • 2. $\text{(a) and (d) only}$
  • 3. $\text{(b) and (c) only}$ (Correct)
  • 4. $\text{(a), (c) and (d) only}$
Solution:
$\text{Hint: } \Delta U = nC_V\Delta T$ $\text{Step 1: Find the change in internal energy.}$ $\text{Given,}$ $T_A = \frac{P_AV_A}{nR}, T_B = \frac{P_BV_B}{nR}$ $\text{In the given diagram all have the same initial and final state. Thus,}$ $P_AV_A = P_BV_B \Rightarrow \frac{P_AV_A}{nR} = \frac{P_BV_B}{nR} \Rightarrow T_A = T_B$ $\text{The change in internal energy for the process } A \text{ to } B\text{, is given by,}$ $\Rightarrow dU_{A\rightarrow B} = nC_VdT = nC_V(T_B - T_A)$ $\text{Since, } T_A = T_B$ $\Rightarrow dU_{A\rightarrow B} = 0$ $\text{Internal energy is a state function and does not depend on path, but on the initial and final state of the gas.}$ $\text{Thus change in internal energy is the same in all four cases.}$ $\text{Step 2: Find the work done.}$ $\text{The work done for } A \text{ to } B\text{, is given by;}$ $\Rightarrow dW_{A\rightarrow B} = \int_A^B PdV$ $\Rightarrow \int_A^B PdV = \text{Area under the } (P-V) \text{ curve from } A \text{ to } B\text{.}$ $\text{It is evident that the area is maximum for the path I.}$ $\text{Thus, the work done is maximum in case I.}$ $\text{Hence, option (3) is the correct answer.}$

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{
  "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
}