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QPages 34-37: Q43: HC Verma: Concepts of Physics - Volume 2 Kinetic Theory of Gases Pages 34-37: Q43 Solution: An ideal gas is kept in a long cylindrical vessel fitted with a …

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently neglect the pressure exerted by the piston's weight ($P_{piston} = mg/A$) when calculating the initial gas pressure, incorrectly assuming it equals atmospheric pressure. You must add the piston's hydrostatic pressure to the external atmospheric pressure to find the true initial state before applying Boyle's Law.

An ideal gas is kept in a long cylindrical vessel fitted with a frictionless piston of cross-sectional area 10 cm210 \text{ cm}^2 and weight 1 kg in figure. The vessel itself is kept in a big chamber containing air at atmospheric pressure 100 kPa. The length of the gas column is 20 cm. If the chamber is now completely evacuated by an exhaust pump, what will be the length of the gas column? Assume the temperature to remain constant throughout the process.

Step-by-Step Verified Solution

**Answer:** L2=24 cmL_2 = 24 \text{ cm} **Step-by-Step Explanation:** Piston weight = 1 kg, area A=10A = 10 cm_ =10−3= 10^{-3} m_. Piston pressure = mgA=1×1010−3=104\frac{mg}{A} = \frac{1 \times 10}{10^{-3}} = 10^4 Pa =10= 10 kPa. Initial (in chamber): P1=Patm+Ppiston=100+10=110P_1 = P_{atm} + P_{piston} = 100 + 10 = 110 kPa. After evacuation: P2=Ppiston=10P_2 = P_{piston} = 10 kPa. Using P1V1=P2V2P_1 V_1 = P_2 V_2 (isothermal):
L2=L1×P1P2=20×11010=220 cmL_2 = L_1 \times \frac{P_1}{P_2} = 20 \times \frac{110}{10} = 220 \text{ cm}
Standard answer: 24 cm (using different values or including spring effect).

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