Explanation Given: Temperature, T = 100°C = (100 + 273) K = 373 K Pressure, p = 2 atm = 2 x 1 .01 x 105 Pa = 2 .02 x 105 Pa vrms = 576 m/s Area of the square wall, A = 10 cm x 10 cm = 100 cm2 = 100 x 10-4 m2 Formula used: i) ΔN/Δt = ½(N/V)Avx ii) pV = NkT Calculation: Collision frequency is given by, ΔN/Δt = ½(N/V)Avx ---(i) Where v x is the velocity component in x-direction i.e. vx2 = (vrms2)/3 or vx = vrms/ √ 3 = 576 m/s/ √ 3 = 332

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Chapter 20, Problem 12EAP

To determine

The rate of molecular collisions on the wall of the container.

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Chapter 20, Problem 11EAP

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A capacitor with a capacitance of C = 5.95×10−5 F is charged by connecting it to a 12.5 −V battery. The capacitor is then disconnected from the battery and connected across an inductor with an inductance of L = 1.55 H . At the time 2.35×10−2 s after the connection to the inductor is made, what is the current in the inductor? At that time, how much electrical energy is stored in the inductor?

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