## Essential University Physics: Volume 1 (3rd Edition)

We find the change in volume of the can and the gas to find: $\Delta V_{can}=\beta V\Delta T=(3)(12\times10^{-6})(20)(15)=.0108L$ $\Delta V_{gas}=\beta V\Delta T=(95\times10^{-5})(20)(15)=.285L$ Thus, .274 Liters are lost.