States of Matter
13479
One atmosphere is numerically equal to approximately
1 \({10^6}\,{\rm{dynes }}\,c{m^{ - 2}}\)
2 \({10^2}\,{\rm{dynes }}\,c{m^{ - 2}}\)
3 \({10^4}\,{\rm{dynes }}\,c{m^{ - 2}}\)
4 \({10^8}\,{\rm{dynes }}\,c{m^{ - 2}}\)
Explanation:
(a) \(1\, atm\) = \({10^6}\) \(dynes\, cm^{-2}\)