Đáp án:
$\begin{align}
& a){{p}_{1}}=8atm \\
& b){{t}_{2}}={{327}^{0}}C \\
& c){{V}_{3}}=27,3lit \\
& d){{V}_{4}}=13,04lit \\
\end{align}$
Giải thích các bước giải:
${{V}_{0}}=30l;{{t}_{0}}={{27}^{0}}C;{{p}_{0}}=2atm$
a) ném đẳng nhiệt
${{t}_{1}}={{t}_{0}}={{27}^{0}}C;{{V}_{1}}=10l$
ta có:
$\begin{align}
& {{p}_{0}}.{{V}_{0}}={{p}_{1}}.{{V}_{1}} \\
& \Leftrightarrow 2.30={{p}_{1}}.10 \\
& \Rightarrow {{p}_{1}}=6atm \\
\end{align}$
b) đẳng tích ${{V}_{0}}={{V}_{2}}=30l;{{p}_{2}}=4atm$
ta có:
$\begin{align}
& \dfrac{{{p}_{0}}}{{{T}_{0}}}=\dfrac{{{p}_{2}}}{{{T}_{2}}} \\
& \Leftrightarrow \dfrac{2}{27+273}=\dfrac{4}{{{t}_{2}}+273} \\
& \Rightarrow {{t}_{2}}={{327}^{0}}C \\
\end{align}$
c) đẳng áp${{p}_{0}}={{p}_{3}}=2atm;{{t}_{3}}={{0}^{0}}C$
ta có:
$\begin{align}
& \frac{{{V}_{0}}}{{{T}_{0}}}=\dfrac{{{V}_{3}}}{{{T}_{3}}} \\
& \Leftrightarrow \dfrac{30}{27+273}=\dfrac{{{V}_{3}}}{273} \\
& \Rightarrow {{V}_{3}}=27,3lit \\
\end{align}$
d) ${{t}_{4}}={{53}^{0}}C;{{p}_{4}}=5atm$
phương trình trạng thái khí lý tưởng;
$\begin{align}
& \dfrac{{{p}_{0}}{{V}_{0}}}{{{T}_{0}}}=\dfrac{{{p}_{4}}.{{V}_{4}}}{{{T}_{4}}} \\
& \Leftrightarrow \dfrac{2.30}{27+273}=\dfrac{5.{{V}_{4}}}{53+273} \\
& \Rightarrow {{V}_{4}}=13,04lit \\
\end{align}$
