Giải \(\eqalign{& V = 3l = {3.10^{ - 3}}({m^3});\cr&p = 200(kPa) = {200.10^3}(Pa); \cr
& T = 16 + 273 = 289(K) \cr
& \mu = {{mRT} \over {pV}} = {{11.8,31.289} \over {{{200.10}^3}{{.3.10}^{ - 3}}}} \approx 44;\left[ {{\mu _{C{O_2}}} = 44g} \right] \cr} \)
