## Essential University Physics: Volume 1 (4th Edition) Clone

a) 1.49 mm b) $1.07\times10^{-7}\ J$
a) Atmospheric pressure is 760 mmHg. Thus, we find that these pressures, which are measured in terms of how much more than atmospheric pressure they are, can be given by: $P_0=840 \ mmHg$ $P_f=885 \ mmHg$ Since volume and pressure are inversely related, it follows: $d =2 \sqrt[3] {\frac{840(.76^3)}{885}}=1.49 \ mm$ b) We know the following equation for the work done: $W = P_1(\frac{4}{3}\pi r^3)ln(\frac{P_1}{P_2})$ Thus, converting 840 mmHg to 111,963 Pa, it follows: $W = (111,963)(\frac{4}{3}\pi (.00076)^3)ln(\frac{885}{840})=1.07\times10^{-7}\ J$