Đáp án:

\(\begin{array}{l}a)2,58kg/{m^3}\\b)10kPa\end{array}\)

Giải thích các bước giải:

a)

Trạng thái 1: \(\left\{ \begin{array}{l}{\rho _1} = 1,29kg/{m^3}\\{p_1} = 1atm\\{V_1}\end{array} \right.\)

Trạng thái 2: \(\left\{ \begin{array}{l}{\rho _2} = ?\\{p_2} = 2atm\\{V_2}\end{array} \right.\)

Quá trình đẳng nhiệt \( \Rightarrow {p_1}{V_1} = {p_2}{V_2}\)

Ta có: \(\left\{ \begin{array}{l}{\rho _2} = \dfrac{m}{{{V_2}}}\\{\rho _1} = \dfrac{m}{{{V_1}}}\end{array} \right.\)

Ta suy ra:

\(\begin{array}{l}\dfrac{{{\rho _2}}}{{{\rho _1}}} = \dfrac{{{V_1}}}{{{V_2}}} = \dfrac{{{p_2}}}{{{p_1}}} = 2\\ \Rightarrow {\rho _2} = 2{\rho _1} = 2.1,29 = 2,58kg/{m^3}\end{array}\)

b)

\(\begin{array}{l}TT1\left\{ \begin{array}{l}{V_1} = 24l\\{p_1}\end{array} \right.\\TT2\left\{ \begin{array}{l}{V_2} = 6l\\{p_2} = {p_1} + \Delta p\end{array} \right.\end{array}\)

Quá trình đẳng nhiệt, ta có: \({p_1}{V_1} = {p_2}{V_2}\)

\(\begin{array}{l} \Leftrightarrow {p_1}.24 = \left( {{p_1} + 30} \right).6\\ \Rightarrow {p_1} = 10kPa\end{array}\)