Đáp án:

 V1=9l

P1=4.10^5Pa

Giải thích các bước giải:
\[{P_2} = {2.10^5} + {P_1};{V_2} = {V_1} - 3l;{P_2}' = {5.10^5} + {P_1};{V_2}' = {V_1} - 5l\]

Áp dụng quá trình đẳng nhiệt:
\({V_1}{P_1} = {V_2}.{P_2} = {V_2}'.{P_2}' \Leftrightarrow \left\{ \begin{array}{l}
({V_1} - 3).({2.10^5} + {P_1}) = {V_1}{P_1}\\
({V_1} - 5).)({5.10^5} + {P_1}) = {V_1}{P_1}
\end{array} \right.\)

thay vào: 
\(\left\{ \begin{array}{l}
({V_1} - 3).({2.10^5} + {P_1}) = {V_1}{P_1}\\
({V_1} - 5).({5.10^5} + {P_1}) = {V_1}{P_1}
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
{2.10^5}{V_1} + {V_1}{P_1} - {6.10^5} - 3{P_1} = {V_1}{P_1}\\
{5.10^5}{V_1} + {V_1}{P_1} - {25.10^5} - 5{P_1} = {V_1}{P_1}
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
{2.10^5}{V_1} - 3{P_1} = {6.10^5}\\
{5.10^5}{V_1} - 5{P_0} = {25.10^5}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{P_1} = {4.10^5}Pa\\
{V_1} = 9l
\end{array} \right.\)