Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 15 - The Laws of Thermodynamics - General Problems - Page 441: 72

Answer

$203\;\rm J$

Work Step by Step

We know that the piston moves without friction. And we know that before heating the piston was at rest and after heating, it moves slowly upward for 1 cm and then stopped. This means that the pressure of gas inside the cylinder is given by $$P_{gas}=P_{a}+P_{piston}$$ whereas $P_{piston}$ is the pressure due to the piston's weight, and $P_a$ is the atmospheric pressure. Thus, $$P_{gas}=P_{a}+\dfrac{mg}{A}$$ Plugging the known; $$P_{gas}=(1.013\times10^5)+\dfrac{0.15\cdot 9.8}{0.08}=1.013\times10^5\;\rm Pa$$ It is obvious now that the pressure occurred but the weight of the piston is negligible. We are heating the gas under constant pressure since the piston is free to move up or down without friction, so we increase the volume of the gas through the heating process. So the piston moves up. So, $$\Delta U=\frac{3}{2}P_{gas}\Delta V=Q-W$$ Solving for $Q$; $$Q=\frac{3}{2}P_{gas}\Delta V+W$$ and the work done by the gas is given by $$W= P_{gas}\Delta V$$ $$Q=\frac{3}{2}P_{gas}\Delta V+P_{gas}\Delta V=2.5P_{gas}\Delta V$$ $$Q= 2.5P_{gas}\Delta V=2.5P_{gas}A\Delta h$$ Plugging the known; $$Q= 2.5\cdot (1.013\times10^5)\cdot 0.080\cdot 1\times10^{-2}$$ $$Q=\color{red}{\bf203}\;\rm J$$
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