Đáp án:

$\begin{align}
  & 1){{V}_{1}}=7,5c{{m}^{3}} \\ 
 & 2) \\ 
 & a){{p}_{2}}=1520mmHg \\ 
 & b){{t}_{3}}=-131,{{6}^{0}}C \\ 
\end{align}$

Giải thích các bước giải:

$\Delta V=15c{{m}^{3}};{{p}_{2}}=\frac{{{p}_{1}}}{3};$

đẳng nhiệt

$\begin{align}
  & {{p}_{1}}{{V}_{1}}={{p}_{2}}{{V}_{2}} \\ 
 & \Leftrightarrow {{p}_{1}}.{{V}_{1}}=\dfrac{{{p}_{1}}}{3}.({{V}_{1}}+15) \\ 
 & \Rightarrow {{V}_{1}}=7,5c{{m}^{3}} \\ 
\end{align}$

Bài 2:

${{V}_{1}}=56lit;{{p}_{1}}=760mmHg;{{T}_{1}}=30+273=303K$

a) ${{T}_{2}}=3{{T}_{1}};{{V}_{2}}=1,5{{V}_{1}}$

Phương trình trạng thái khí lí tưởng

$\begin{align}
  & \dfrac{{{p}_{1}}{{V}_{1}}}{{{T}_{1}}}=\dfrac{{{p}_{2}}{{V}_{2}}}{{{T}_{2}}} \\ 
 & \Leftrightarrow \dfrac{760.{{V}_{1}}}{{{T}_{1}}}=\dfrac{{{p}_{2}}.1,5{{V}_{1}}}{3{{T}_{1}}} \\ 
 & \Rightarrow {{p}_{2}}=1520mmHg \\ 
\end{align}$

b) ${{V}_{3}}=\dfrac{{{V}_{1}}}{3};{{p}_{3}}=1,4{{p}_{1}}$

phương trình trạng thái:

$\begin{align}
  & \dfrac{{{p}_{1}}{{V}_{1}}}{{{T}_{1}}}=\dfrac{{{p}_{2}}{{V}_{2}}}{{{T}_{2}}} \\ 
 & \Leftrightarrow \dfrac{{{p}_{1}}.{{V}_{1}}}{303}=\dfrac{1,4{{p}_{1}}.\frac{{{V}_{1}}}{3}}{{{T}_{3}}} \\ 
 & \Rightarrow {{T}_{3}}=141,4K \\ 
 & \Rightarrow {{t}_{3}}=-131,{{6}^{0}}C \\ 
\end{align}$