Đáp án:
$\begin{align}
& a){{T}_{1}}=300K;{{p}_{1}}=1atm;{{V}_{1}}=0,025{{m}^{3}} \\
& {{T}_{2}}=300K;{{p}_{2}}=3atm;{{V}_{2}}=\frac{1}{120}{{m}^{3}} \\
& {{T}_{3}}=900K;{{p}_{3}}=3atm;{{V}_{3}}=0,025{{m}^{3}} \\
& b)\Delta {{U}_{12}}=0J \\
& \Delta {{U}_{23}}=14840J \\
& \Delta {{U}_{31}}=Q \\
\end{align}$
Giải thích các bước giải:
$\begin{align}
& {{T}_{1}}={{T}_{2}}=300K;{{p}_{1}}=1atm \\
& {{T}_{3}}=900K \\
& R=8,31J/mol.K; \\
\end{align}$
a) (1)-> (2) đẳng nhiệt
$\begin{align}
& {{p}_{1}}.{{V}_{1}}=n.R.{{T}_{1}} \\
& \Rightarrow {{V}_{1}}=\dfrac{1.8,31.300}{1,{{013.10}^{5}}}=0,0025{{m}^{3}} \\
\end{align}$
(3)-> (1) đẳng tích
$\begin{align}
& \frac{{{p}_{3}}}{{{T}_{3}}}=\dfrac{{{p}_{1}}}{{{T}_{1}}} \\
& \Rightarrow \frac{{{p}_{3}}}{900}=\dfrac{1}{300} \\
& \Rightarrow {{p}_{3}}=3atm \\
\end{align}$
(2)->(3) đẳng áp
$\begin{align}
& \dfrac{{{V}_{2}}}{{{T}_{2}}}=\dfrac{{{V}_{3}}}{{{T}_{3}}} \\
& \Leftrightarrow \dfrac{{{V}_{2}}}{300}=\dfrac{0,025}{900} \\
& \Rightarrow {{V}_{2}}=\dfrac{1}{120}{{m}^{3}} \\
\end{align}$
b) Độ biến thiên nội năng
$\begin{align}
& \Delta {{U}_{12}}=0J \\
& \Delta {{U}_{23}}=Q-A=19905-{{p}_{2}}.\Delta {{V}_{23}} \\
& =19905-3.1,{{013.10}^{5}}.(0,025-\frac{1}{120}) \\
& =14840J \\
& \Delta {{U}_{31}}=Q=14840J \\
\end{align}$
