Đáp án:
$\begin{array}{l}
a.293K\\
b.2,85l\\
c.4,{5.10^5}Pa
\end{array}$
Giải thích các bước giải:
$\begin{array}{l}
a.\frac{{{P_1}{V_1}}}{{{T_1}}} = \frac{{{P_2}{V_2}}}{{{T_2}}}\\
\Rightarrow \frac{{{{2.10}^5}.4}}{{20 + 273}} = \frac{{{{10}^5}.8}}{{{T_2}}}\\
\Rightarrow {T_2} = 293K\\
b.\frac{{{P_1}{V_1}}}{{{T_1}}} = \frac{{{P_2}{V_2}}}{{{T_2}}}\\
\Rightarrow \frac{{{{2.10}^5}.4}}{{20 + 273}} = \frac{{{{3.10}^5}.{V_2}}}{{40 + 273}}\\
\Rightarrow {V_2} = 2,85l\\
c.\frac{{{P_1}{V_1}}}{{{T_1}}} = \frac{{{P_2}{V_2}}}{{{T_2}}}\\
\Rightarrow \frac{{{{2.10}^5}.4}}{{20 + 273}} = \frac{{{P_2}.2}}{{60 + 273}}\\
\Rightarrow {P_2} = 4,{5.10^5}Pa
\end{array}$