Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 14 - Partial Derivatives - 14.4 Exercises - Page 947: 38

Answer

$-8.83 k Pa$

Work Step by Step

The differential form can be evaluated as follows: $dP=\dfrac{\partial P}{\partial V} dV + \dfrac{\partial P}{\partial T} dT$ Take the partial derivatives. $dP=\dfrac{-8.31 T}{V^2} dV + \dfrac{8.31}{V} dT$ Re-arrange as: $\triangle P \approx \dfrac{-8.31 T}{V^2} \triangle V + \dfrac{8.31 T}{V} \triangle T$ Plug in the given data and we have $\triangle P \approx \dfrac{-8.31 \times 310}{(12)^2} \cdot (0.3) + \dfrac{8.31}{12} \cdot (-5)$ Hence, we have $\triangle P \approx -8.83 k Pa$
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