(c) $\mathrm{\Delta}p = 100\ atm = 100\ \times 10^{6}\ dyne\ cm^{2}$ = $10^{8}\ dyne\ \ cm^{- 2}\ $ $\frac{\mathrm{\Delta}V}{V} = \frac{0.01}{100} = 10^{- 4}\ $ $K = \frac{10^{8}}{10^{- 4}}\ dyne\ cm^{- 2} = 10^{12}\ dyne\ cm^{- 2}$
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Ph.D. in Physics with 19+ years of teaching experience.