Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 16 - A Macroscopic Description of Matter - Exercises and Problems - Page 465: 39

Answer

$107^\circ \rm C$

Work Step by Step

We know, for an ideal gas, that $$PV=nRT$$ So, the temperature of the gas is given by $$T=\dfrac{PV}{nR}\tag 1$$ Now we need to find the number of moles of oxygen which is given by $$n=\dfrac{m}{M}=\dfrac{m}{2M_O}$$ where $m$ is the mass of the oxygen inside the cylinder and $M_O$ is the atomic mass of the oxygen element. Noting that the oxygen gas consists of two oxygen atoms hence the 2 in front of $M_O$. Plugging the known; $$n=\dfrac{100\times 10^{-3}\;\rm g}{2(16\;\rm g/mol)}$$ $$n=\bf 3.125\times 10^{-3}\;\rm mol\tag 2$$ Now we know that the pressure inside the cylinder is given by $$P=P_{out}-P_{pull}$$ $$P=P_{out}-\dfrac{F_{pull}}{A}=P_{out}-\dfrac{F_{pull}}{\pi r^2}$$ Plugging the known; $$P=(100\times 10^3)-\dfrac{(184)}{\pi (3\times 10^{-2})^2}$$ $$P=\bf 3.49\times 10^4\;\rm Pa\tag 3$$ The volume of the gas is the volume of the cylinder which is given by $$V=Ah=\pi r^2 h$$ Plugging the known; $$V =\pi (3\times 10^{-2})^2 (10\times 10^{-2})$$ $$V=\bf 2.83\times 10^{-4}\;\rm m^3\tag 4$$ Plugging the known and from (2), (3), and (4) into (1); $$T=\dfrac{(3.49\times 10^4)(2.83\times 10^{-4})}{(3.125\times 10^{-3})(8.31)}\approx \bf 380\;\rm K$$ Hence, $$T=380-273=\color{red}{\bf 107^\circ}\rm C$$
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