Đáp án:

Câu 1:

Ta có:

\[\frac{{{p_1}}}{{{p_2}}} = \frac{{{T_1}}}{{{T_2}}} \Rightarrow \left\{ \begin{array}{l}
{p_1} = \frac{{{T_1}}}{{{T_2}}}{p_2}\\
{p_2} = \frac{{{T_2}}}{{{T_1}}}{p_1}\\
{T_1} = \frac{{{p_1}}}{{{p_2}}}{T_2}\\
{T_2} = \frac{{{p_2}}}{{{T_2}}}{T_1}
\end{array} \right.\]

Câu 2:

Ta có:

\[\frac{{{V_1}}}{{{V_2}}} = \frac{{{T_1}}}{{{T_2}}} \Rightarrow \left\{ \begin{array}{l}
{V_1} = \frac{{{T_1}}}{{{T_2}}}{V_2}\\
{V_2} = \frac{{{T_2}}}{{{T_1}}}{V_1}\\
{T_1} = \frac{{{V_1}}}{{{V_2}}}{T_2}\\
{T_2} = \frac{{{V_2}}}{{{V_1}}}{T_1}
\end{array} \right.\]

Câu 3:

Áp dụng phương trình đẳng tích ta có:

\[\frac{{{p_1}}}{{{p_2}}} = \frac{{{T_1}}}{{{T_2}}} \Rightarrow {p_2} = \frac{{{T_2}}}{{{T_1}}}{p_1} = \frac{{40 + 273}}{{20 + 273}}.1,{5.10^5} = 1,{6.10^5}Pa\]