Đáp án:
$\begin{align}
& {{V}_{1}}=2lit;{{P}_{1}}=1,5atm;{{T}_{1}}=300k \\
& {{V}_{2}}=1,5lit;{{P}_{2}}=2atm;{{T}_{2}}=300k \\
& {{V}_{3}}=1,5lit;{{P}_{3}}=4atm;{{T}_{3}}=800k \\
& {{V}_{4}}=1,125lit;{{P}_{4}}=4atm;{{T}_{4}}=600K \\
\end{align}$
Giải thích các bước giải:
${{P}_{1}}=1,5atm;{{V}_{1}}=2lit;{{T}_{1}}=300K;$
a) ${{V}_{2}}=1,5lit;{{P}_{2}}=?$
quá trình đẳng nhiệt:
$\begin{align}
& {{P}_{1}}{{V}_{1}}={{P}_{2}}.{{V}_{2}} \\
& \Rightarrow {{P}_{2}}={{P}_{1}}.\frac{{{V}_{1}}}{{{V}_{2}}}=1,5.\frac{2}{1,5}=2atm \\
\end{align}$
b)${{P}_{3}}=2{{P}_{2}}=4atm;$
đẳng tích:
$\begin{align}
& \frac{{{P}_{2}}}{{{T}_{2}}}=\frac{{{P}_{3}}}{{{T}_{3}}} \\
& \Rightarrow {{T}_{3}}={{T}_{2}}.\frac{{{P}_{3}}}{{{P}_{2}}}=300.\frac{4}{1,5}=800K \\
\end{align}$
c) đẳng áp
${{P}_{4}}={{P}_{3}}=4atm;{{T}_{4}}=600K$
$\begin{align}
& \frac{{{V}_{3}}}{{{T}_{3}}}=\dfrac{{{V}_{4}}}{{{T}_{4}}} \\
& \Rightarrow {{V}_{4}}={{V}_{3}}.\dfrac{{{T}_{4}}}{{{T}_{3}}}=1,5.\dfrac{600}{800}=1,125lit \\
\end{align}$