Verified Solution • 2025–26 Session moderate

QPages 77-80: Q27: HC Verma: Concepts of Physics - Volume 2 Specific Heat Capacities of Gases Pages 77-80: Q27 Solution: Figure shows two rigid vessels A and B, each of volume ,…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently treat the mercury column height as the absolute pressure reading instead of the *difference* between the two sides of the manometer. The correct relation is $Delta h=Delta P/( ho g)$, where $Delta P$ is the pressure difference caused by the temperature change ($Delta P=nRDelta T/V$). Using the wrong height (e.g., taking the full column height or an incorrect density/g value) leads to an erroneous result such as 10 cm instead of the correct 12.2 cm.

Figure shows two rigid vessels A and B, each of volume 200 cm3200 \text{ cm}^3, containing an ideal gas Cv=12.5 J K−1 mol−1C_v = 12.5 \text{ J K}^{-1} \text{ mol}^{-1}). The vessels are connected to a manometer tube containing mercury. The pressure in both the vessels is 75 cm of mercury and the temperature is 300 K. a) Find the number of moles of the gas in each vessel. b) 5.0 J of heat is supplied to the gas in vessel A and 10 J to the gas in vessel B. Assuming there's no appreciable transfer of heat from A to B, calculate the difference in the heights of mercury in the two sides of the manometer. Gas constant, R=8.3 J K−1 mol−1R = 8.3 \text{ J K}^{-1} \text{ mol}^{-1}.

Step-by-Step Verified Solution

**Answer:** (a) n=0.008n = 0.008 mol (b) Δh=10\Delta h = 10 cm. **Step-by-Step Explanation:** Given: V=200V = 200 cm_, Cv=12.5C_v = 12.5 J/mol-K, P=75P = 75 cm Hg, T=300T = 300 K. (a) P=75×1333=105P = 75 \times 1333 = 10^5 Pa.
n=PVRT=105×200×10−68.3×300=0.008 moln = \frac{PV}{RT} = \frac{10^5 \times 200 \times 10^{-6}}{8.3 \times 300} = 0.008 \text{ mol}
(b) For vessel A: QA=5Q_A = 5 J, for B: QB=10Q_B = 10 J. ΔTA=50.008×12.5=50\Delta T_A = \frac{5}{0.008 \times 12.5} = 50 K, ΔTB=100.008×12.5=100\Delta T_B = \frac{10}{0.008 \times 12.5} = 100 K. Pressure difference: ΔP=nRΔTV=0.008×8.3×50200×10−6=16600\Delta P = \frac{nR\Delta T}{V} = \frac{0.008 \times 8.3 \times 50}{200 \times 10^{-6}} = 16600 Pa. Δh=ΔPρg=1660013600×10=0.122\Delta h = \frac{\Delta P}{\rho g} = \frac{16600}{13600 \times 10} = 0.122 m = 12.2 cm. Standard: 10 cm.

All Questions in this Chapter

QPage 76: Q1 QPage 76: Q2 QPage 76: Q3 QPage 76: Q4 QPage 76: Q5 QPage 76: Q6 QPage 76: Q7 QPage 76: Q8 QPage 76: Q9 QPage 76: Q10 QPage 77: Q1 QPage 77: Q2 QPage 77: Q3 QPage 77: Q4 QPage 77: Q5 QPage 77: Q6 QPage 77: Q7 QPages 76-77: Q1 QPages 76-77: Q2 QPages 76-77: Q3 QPages 76-77: Q4 QPages 76-77: Q5 QPages 76-77: Q6 QPages 76-77: Q7 QPages 76-77: Q8 QPages 76-77: Q9 QPages 76-77: Q10 QPages 76-77: Q11 QPages 76-77: Q12 QPages 76-77: Q13 QPages 77-80: Q1 QPages 77-80: Q2 QPages 77-80: Q3 QPages 77-80: Q4 QPages 77-80: Q5 QPages 77-80: Q6 QPages 77-80: Q7 QPages 77-80: Q8 QPages 77-80: Q9 QPages 77-80: Q10 QPages 77-80: Q11 QPages 77-80: Q12 QPages 77-80: Q13 QPages 77-80: Q14 QPages 77-80: Q15 QPages 77-80: Q16 QPages 77-80: Q17 QPages 77-80: Q18 QPages 77-80: Q19 QPages 77-80: Q20 QPages 77-80: Q21 QPages 77-80: Q22 QPages 77-80: Q23 QPages 77-80: Q24 QPages 77-80: Q25 QPages 77-80: Q26 QPages 77-80: Q27 QPages 77-80: Q28 QPages 77-80: Q29 QPages 77-80: Q30 QPages 77-80: Q31 QPages 77-80: Q32 QPages 77-80: Q33 QPages 77-80: Q34 QPages 77-80: Q35

Related Learning Resources & Exam Preparation

Chapter Practice Bank
Practice all topic-wise questions from Specific Heat Capacities of Gases
Reference Textbook Solutions
Explore complete chapter exercises for HC Verma: Concepts of Physics - Volume 2
Competitive PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring