Giải bài 2 trang 237 – Bài 48 - SGK môn Vật lý lớp 10 Nâng cao

$\begin{align} & V=10\text{ }l\text{ =1}{{\text{0}}^{-2}}\text{ }\left( {{m}^{3}} \right);\text{ p = 250 }\left( kPa \right)\text{ = 2,5}\text{.1}{{\text{0}}^{5}}\text{ }\left( Pa \right); \\ & T=27+273=300\text{ }\left( K \right);\text{ }{{\mu }_{{{O}_{2}}}}=32\text{ }\left( g \right) \\ & m=\dfrac{\mu pV}{RT}={{\dfrac{32.2,{{5.10}^{5}}.10}{8,31.3000}}^{-2}}\approx 32,1\text{ }\left( g \right). \\ \end{align}$ 

Ghi nhớ:

Phương trình Cla-pê-rôn-Men-đê-lê-ép:

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