First Law Of Thermodynamics
A gas in a cylinder absorbs 500 J of heat while the gas does 200 J of work in pushing a piston outward, so by how much does its internal energy change?
Select the correct option:
Solution
300 J
According to NCERT Class 11, Chapter 12 (Thermodynamics), the First Law of Thermodynamics is a statement of energy conservation and is written as ΔQ=Δcup+ΔW, where ΔQ is heat supplied to the gas, ΔU is the change in internal energy, and ΔW is the work done by the gas. Rearranging gives Δcup=ΔQ−ΔW. Substituting the given values: Δcup=500−200=300 J. Because the system absorbs heat (ΔQ positive) and does work on the surroundings (ΔW positive), part of the supplied heat leaves as work and the remainder raises the internal energy. The option 700 J is wrong because it incorrectly adds the work instead of subtracting it. The option −300 J is wrong as it reverses the sign convention, treating work done by the gas as added to internal energy with the wrong sign. The option 100 J is wrong because it does not follow from the correct subtraction. A plausibility check: ΔU must be less than ΔQ whenever the gas does positive work, and 300 J < 500 J confirms the sign and magnitude are reasonable.
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About This Question
- Subject
- physics
- Chapter
- thermodynamics
- Topic
- first law of thermodynamics
- Difficulty
- Easy
- Year
- 2025
Solution
Correct Answer:
300 J
According to NCERT Class 11, Chapter 12 (Thermodynamics), the First Law of Thermodynamics is a statement of energy conservation and is written as ΔQ=Δcup+ΔW, where ΔQ is heat supplied to the gas, ΔU is the change in internal energy, and ΔW is the work done by the gas. Rearranging gives Δcup=ΔQ−ΔW. Substituting the given values: Δcup=500−200=300 J. Because the system absorbs heat (ΔQ positive) and does work on the surroundings (ΔW positive), part of the supplied heat leaves as work and the remainder raises the internal energy. The option 700 J is wrong because it incorrectly adds the work instead of subtracting it. The option −300 J is wrong as it reverses the sign convention, treating work done by the gas as added to internal energy with the wrong sign. The option 100 J is wrong because it does not follow from the correct subtraction. A plausibility check: ΔU must be less than ΔQ whenever the gas does positive work, and 300 J < 500 J confirms the sign and magnitude are reasonable.
This easy difficulty physics question is from the chapter thermodynamics, covering the topic of first law of thermodynamics. It appeared in the 2025 exam.
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