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MCQEasyJEE 2023First Law & Internal Energy

JEE Physics 2023 Question with Solution

1 g of a liquid is converted to vapour at 3 × 10⁵ Pa pressure. If 10% of the heat supplied is used for increasing the volume by 1600 cm³ during this phase change, then the increase in internal energy in the process will be :

  • A

    4320 J

  • B

    432000 J

  • C

    4800 J

  • D

    4.32 × 10⁸ J

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: Pressure P=3×105 PaP = 3 \times 10^5 \, \text{Pa} and increase in volume ΔV=1600×10−6 m3\Delta V = 1600 \times 10^{-6} \, \text{m}^3. Also, 10%10\% of the heat supplied is used in expansion work.

Find: Increase in internal energy.

Work done is

W=PΔV=3×105×1600×10−6W = P\Delta V = 3 \times 10^5 \times 1600 \times 10^{-6} W=480 JW = 480 \, \text{J}

Only 10%10\% of the heat is used in doing work, so total heat supplied is

Q=4800.1=4800 JQ = \frac{480}{0.1} = 4800 \, \text{J}

Therefore, the remaining 90%90\% goes into internal energy:

ΔU=0.9×4800=4320 J\Delta U = 0.9 \times 4800 = 4320 \, \text{J}

Therefore, the increase in internal energy is 4320 J4320 \, \text{J}. The solution states option B, but this value matches option A in the listed options.

Percentage Split

Given: Expansion work is 480 J480 \, \text{J}, and this is 10%10\% of the supplied heat.

Find: Internal energy increase.

If 10%10\% of heat equals 480 J480 \, \text{J}, then 90%90\% of heat is nine times this amount:

ΔU=9×480=4320 J\Delta U = 9 \times 480 = 4320 \, \text{J}

Therefore, the correct option is A.

Common mistakes

  • Using 1600 cm31600 \, \text{cm}^3 directly in PΔVP\Delta V without converting it to m3\text{m}^3 is incorrect because SI units are required with pressure in pascal. Convert 1600 cm31600 \, \text{cm}^3 to 1600×10−6 m31600 \times 10^{-6} \, \text{m}^3 first.

  • Taking 480 J480 \, \text{J} as the internal energy change is wrong because the question states that this part is only the expansion work, which is 10%10\% of the total heat supplied. First find total heat, then take the remaining 90%90\% for internal energy.

  • Assuming the answer must be option B because the solution labels it so is incorrect here. The worked value is 4320 J4320 \, \text{J}, which matches option A in the given options, so the numerical working should be trusted over the mislabeled option letter.

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