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Current Question (ID: 8289)

Question:

$\text{701 J of heat is absorbed by a system and 394 J of work is done by the system. The change in internal energy for the process is:}$

Options:

1. $307 \text{ J}$
2. $-307 \text{ J}$
3. $1095 \text{ J}$
4. $-701 \text{ J}$

Solution:

$\text{Hint: First Law of Thermodynamics}$ $\text{Step 1: Apply the First Law of Thermodynamics}$ $\text{According to the first law of thermodynamics,}$ $\Delta U = q + W \quad (i)$ $\text{Where,}$ $\Delta U = \text{change in internal energy for a process}; q = \text{heat}; W = \text{work}$ $\text{Step 2: Determine the sign conventions and substitute the values}$ $\text{Given,}$ $q = +701 \text{ J (Since heat is absorbed by the system, q is positive)}$ $W = -394 \text{ J (Since work is done by the system on the surroundings, W is negative)}$ $\text{In thermodynamics, the sign conventions are:}$ $\text{• Heat absorbed by the system (q) is positive}$ $\text{• Heat released by the system (q) is negative}$ $\text{• Work done on the system (W) is positive}$ $\text{• Work done by the system (W) is negative}$ $\text{Step 3: Calculate the change in internal energy}$ $\text{Substituting the values in expression (i), we get:}$ $\Delta U = 701 \text{ J} + (-394 \text{ J})$ $\Delta U = 701 \text{ J} - 394 \text{ J}$ $\Delta U = 307 \text{ J}$ $\text{The positive value of } \Delta U \text{ indicates that the internal energy of the system has increased.}$ $\text{Hence, the change in internal energy for the given process is 307 J.}$

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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
}