Verified Solution • 2025–26 Session moderate

QPages 34-37: Q26: HC Verma: Concepts of Physics - Volume 2 Kinetic Theory of Gases Pages 34-37: Q26 Solution: Air is pumped into the tubes of a cycle rickshaw at a pressure…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often mistakenly use the given gauge pressure (2 atm) directly in the ideal gas law without converting it to absolute pressure, or they incorrectly assume the final pressure inside the punctured tube is zero. You must add atmospheric pressure ($1 \text{ atm}$) to the gauge pressure to get the initial absolute pressure ($3 \text{ atm}$) and recognize that the final pressure equals atmospheric pressure ($1 \text{ atm}$) for the correct mole calculation.

Air is pumped into the tubes of a cycle rickshaw at a pressure of 2 atm. The volume of each tube at this pressure is

0.002 \text{ m}^3

. One of the tubes gets punctured and the volume of the tube reduces to

0.0005 \text{ m}^3

. How many moles of air have leaked out? Assume that the temperature remains constant at 300 K and that the air behaves as an ideal gas. Use

R = 8.3 \text{ J K}^{-1} \text{ mol}^{-1}

Step-by-Step Verified Solution

**Answer:** Moles leaked

= 0.14

**Step-by-Step Explanation:** Initial:

P_1 = 2 \times 10^5

Pa,

V_1 = 0.002

m_,

T = 300

K.

n_1 = \frac{2 \times 10^5 \times 0.002}{8.3 \times 300} = 0.16 \text{ mol}

Final:

P_2 = 10^5

Pa,

V_2 = 0.0005

m_.

n_2 = \frac{10^5 \times 0.0005}{8.3 \times 300} = 0.02 \text{ mol}

Leaked

= 0.16 - 0.02 = 0.14

mol.

All Questions in this Chapter

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