Giải bài 3 trang 237 – Bài 48 - SGK môn Vật lý lớp 10 Nâng cao

$\begin{align} & V=3\text{ }l\text{ = 3}\text{.1}{{\text{0}}^{-3}}\text{ }\left( {{m}^{3}} \right);\text{ p = 200 }\left( kPa \right)\text{ =200}\text{.1}{{\text{0}}^{3}}\text{ }\left( Pa \right); \\ & T=16+273=289\text{ }\left( K \right) \\ & \mu =\dfrac{mRT}{pV}=\dfrac{11.8,31.289}{{{200.10}^{3}}{{.3.10}^{-3}}}\approx 44\text{ }\left( g \right);\text{ }\left[ {{\mu }_{C{{O}_{2}}}}=44g \right] \\ \end{align}$ 

Ghi nhớ:

Phương trình Cla-pê-rôn-Men-đê-lê-ép:

$pV=vRT=\dfrac{m}{\mu }RT$