Câu 4:

Ta có: 

\(\begin{array}{l}
{p_0}{V_0} = {p_1}{V_1} = {p_2}{V_2}\\
 \Rightarrow \left\{ \begin{array}{l}
{p_0}{V_0} = \left( {{p_0} + {{2.10}^5}} \right)\left( {{V_0} - 3} \right)\\
{p_0}{V_0} = \left( {{p_0} + {{5.10}^5}} \right)\left( {{V_0} - 5} \right)
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
{p_0}{V_0} = {p_0}{V_0} - 3{p_0} + {2.10^5}{V_0} - {6.10^5}\\
{p_0}{V_0} = {p_0}{V_0} - 5{p_0} + {5.10^5}{V_0} - {25.10^5}
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
{p_0} = {4.10^5}Pa\\
{V_0} = 9l
\end{array} \right.
\end{array}\)

Câu 5:

Áp dụng định luật Bôi lơ - Ma ri ốt, ta có:

\({p_0} + \dfrac{h}{{13,6}}{V_1} = {p_0}.1,5{V_1} \Rightarrow h = \dfrac{{13,6}}{2}{p_0} = 510cm = 5,1m\)