Đáp án:

$\begin{align}
  & a)N=1,{{0122.10}^{24}}hat \\ 
 & b){{p}_{2}}=2atm \\ 
 & c){{T}_{3}}=75K \\ 
 & d){{p}_{4}}=8atm \\ 
\end{align}$

Giải thích các bước giải:

${{p}_{1}}=4atm;{{V}_{1}}=2lit;{{T}_{1}}=27+273=300K$

a) phương trình Cla-pê-rôn – Men-đê-lê-ép:

$\begin{align}
  & {{p}_{1}}{{V}_{1}}=n.R.{{T}_{1}} \\ 
 & \Rightarrow n=\frac{{{p}_{1}}{{V}_{1}}}{R.{{T}_{1}}}=\frac{{{4.2.10}^{3}}}{8,31.300}=1,68mol \\ 
\end{align}$

số hạt khí:

$N=n.{{N}_{A}}=1,68.6,{{025.10}^{23}}=1,{{0122.10}^{24}}hat$

b) đẳng nhiệt: ${{T}_{1}}={{T}_{2}}$

$\begin{align}
  & {{p}_{1}}{{V}_{1}}={{p}_{2}}{{V}_{2}} \\ 
 & \Rightarrow {{p}_{2}}=\frac{{{p}_{1}}{{V}_{1}}}{{{V}_{2}}}=\frac{4.2}{4}=2atm \\ 
\end{align}$

c) đẳng áp:${{p}_{2}}={{p}_{3}}$

$\begin{align}
  & \frac{{{V}_{2}}}{{{T}_{2}}}=\dfrac{{{V}_{3}}}{{{T}_{3}}} \\ 
 & \Rightarrow {{T}_{3}}=\dfrac{{{T}_{2}}{{V}_{3}}}{{{V}_{2}}}=\dfrac{300.1}{4}=75K \\ 
\end{align}$

d) đẳng tích: ${{T}_{4}}=27+273=300K$

$\begin{align}
  & \frac{{{p}_{3}}}{{{T}_{3}}}=\dfrac{{{p}_{4}}}{{{T}_{4}}} \\ 
 & \Rightarrow {{p}_{4}}=\dfrac{{{T}_{4}}{{p}_{3}}}{{{T}_{3}}}=\dfrac{300.2}{75}=8atm \\ 
\end{align}$