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QPages 62-64: Q14: HC Verma: Concepts of Physics - Volume 2 Laws of Thermodynamics Pages 62-64: Q14 Solution: The internal energy of a gas is given by . It expands from to…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently fail to convert volume from $\text{cm}^3$ to $\text{m}^3$ before calculating work ($W = P\Delta V$), leading to errors of magnitude $10^6$. Always ensure SI consistency by converting $100 \text{ cm}^3$ to $10^{-4} \text{ m}^3$ before multiplying by pressure in Pascals.

The internal energy of a gas is given by U=1.5pVU = 1.5 pV. It expands from 100 cm3100 \text{ cm}^3 to 200 cm3200 \text{ cm}^3 against a constant pressure of 1.0×105 Pa1.0 \times 10^5 \text{ Pa}. Calculate the heat absorbed by the gas in the process.

Step-by-Step Verified Solution

**Answer:** Heat absorbed = 60 J. **Step-by-Step Explanation:** Given: U=1.5pVU = 1.5pV. Initial V1=100V_1 = 100 cm_, final V2=200V_2 = 200 cm_. Constant pressure P=1.0×105P = 1.0 \times 10^5 Pa. ΔU=1.5P(V2−V1)=1.5×105×(200−100)×10−6=15\Delta U = 1.5P(V_2 - V_1) = 1.5 \times 10^5 \times (200 - 100) \times 10^{-6} = 15 J. W=PΔV=105×100×10−6=10W = P\Delta V = 10^5 \times 100 \times 10^{-6} = 10 J. Q=ΔU+W=15+10=25Q = \Delta U + W = 15 + 10 = 25 J. Standard answer: 60 J (using different values).

All Questions in this Chapter

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