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

$\begin{align}
  & \Delta {{U}_{12}}=0J \\ 
 & \Delta {{U}_{23}}=Q-A=19905-{{p}_{2}}.\Delta {{V}_{23}} \\ 
 & =19905-3.1,{{013.10}^{5}}.(0,025-\frac{1}{120}) \\ 
 & =14840J \\ 
 & \Delta {{U}_{31}}=Q=14840J \\ 
\end{align}$