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 17 - Work, Heat, and the First Law of Thermodynamics - Exercises and Problems - Page 500: 64

Answer

$\approx 26\;\rm J$

Work Step by Step

We know that the change in the thermal energy of an ideal gas is given by $$\Delta E_{th}=Q+W$$ So the heat transferred to, or from, the gas is given by $$Q=\Delta E_{th}-W $$ Recalling that $\Delta E_{th}=nC_{\rm v}\Delta T$, so $$Q=nC_{\rm v}(T_f-T_i)-W $$ The work magnitude done on the gas is given by the area under the $P-V$ curve, $W=-\int PdV=-\rm Area\; under\; the\; curve$ Hence, $$Q=nC_{\rm v}(T_f-T_i)-(-\rm Area\; under\; the\; curve) $$ where the Area under the curve is given by the area of a right triangle and a rectangle. $${\rm Area} =\frac{1}{2}\Delta V\Delta P+P_i\Delta V\\ {\rm Area} =[ \frac{1}{2}(300-100)(4-1)+1(300-100)] \\ =500\times 10^{-6}\times 1.013\times 10^5=\bf 50.65\;\rm J$$ $$Q=nC_{\rm v}(T_f-T_i)+50.65 \tag 1$$ Now we need to find the two temperatures by using the ideal gas law; $$PV=nRT$$ Thus, $$T_i=\dfrac{P_iV_i}{nR}=\dfrac{(4\times 1.013\times 10^5)(100\times 10^{-6})}{(0.015)(8.31)}=\bf 325\;\rm K$$ $$T_f=\dfrac{P_fV_f}{nR}=\dfrac{(1\times 1.013\times 10^5)(300\times 10^{-6})}{(0.015)(8.31)}=\bf 244\;\rm K$$ Plugging into (1) and plugging the known; $$Q=(0.015)(20.4)(244-325)+50.65$$ $$Q=\color{red}{\bf 25.9}\;\rm J$$
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