# Chapter 18 - Exercises and Problems - Page 331: 36

$-6689 \ J$

#### Work Step by Step

We know the following equation for the work it will take: $W_2=\frac{ln(c_2)}{ln(c_1)}(W_1)$ Where $c_1$ is the fraction that the volume is initially changed by and where $c_2$ is the factor by which the volume will be changed by. Thus, we find: $W_2=\frac{ln(\frac{1}{22})}{ln(\frac{1}{2})}(-1500)=-6689 \ J$

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