## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson

# Chapter 18 - A Macroscopic Description of Matter - Exercises and Problems - Page 511: 22

#### Answer

(a) The new volume is $1.27~V_0$ (b) The new volume is $2~V_0$

#### Work Step by Step

We can use the ideal gas law to solve this question. $PV = nRT$ $V = \frac{nRT}{P}$ Let $V_0$ be the initial volume. Let $T_1$ be the initial temperature in Kelvin. $V_0 = \frac{nRT_1}{P}$ (a) We can find the value of $T_1$ in Kelvin. $K = C + 273$ $K = 100 + 273$ $K = 373$ If the Celsius temperature is doubled, we can find the new temperature $T_2$ in Kelvin. $K = C + 273$ $K = 200 + 273$ $K = 473$ We can find an expression for $T_2$ in terms of $T_1$: $\frac{T_2}{T_1} = \frac{473}{373}$ $T_2 = 1.27~T_1$ We can find an expression for the new volume $V_2$: $V_2 = \frac{nRT_2}{P}$ $V_2 = \frac{nR(1.27~T_1)}{P}$ $V_2 = 1.27~\frac{nRT_1}{P}$ $V_2 = 1.27~V_0$ The new volume is $1.27~V_0$. (b) If the Kelvin temperature is doubled, then $T_2 = 2~T_1$ We can find an expression for the new volume $V_2$: $V_2 = \frac{nRT_2}{P}$ $V_2 = \frac{nR(2~T_1)}{P}$ $V_2 = 2~\frac{nRT_1}{P}$ $V_2 = 2~V_0$ The new volume is $2~V_0$.

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