Đáp án:
\(\begin{array}{l}
47.A\\
{T_1} = {t_1} + 273 = 20 + 273 = 293^\circ K\\
{T_2} = {t_2} + 273 = 40 + 273 = 313^\circ K\\
\dfrac{{{p_1}}}{{{T_1}}} = \dfrac{{{p_2}}}{{{T_2}}} \Rightarrow {p_2} = \dfrac{{{p_1}{T_2}}}{{{T_1}}} = \dfrac{{{{10}^5}.313}}{{293}} = 1,{068.10^5}Pa\\
49.D\\
{T_1} = {t_1} + 273 = 27 + 273 = 300^\circ K\\
{V_2} = {V_1} - 1,8 = 2 - 1,8 = 0,2d{m^3}\\
{p_2} = {p_1} + 14 = 1 + 14 = 15atm\\
\dfrac{{{p_1}{V_1}}}{{{T_1}}} = \dfrac{{{p_2}{V_2}}}{{{T_2}}} \Rightarrow {T_2} = \dfrac{{{p_2}{V_2}{T_1}}}{{{p_1}{V_1}}} = \dfrac{{15.0,2.300}}{{2.1}} = 450^\circ K\\
\Rightarrow {t_2} = {T_2} - 273 = 450 - 273 = 177^\circ K\\
51.C\\
{T_1} = {t_1} + 273 = 27 + 273 = 300^\circ K\\
{T_2} = {t_2} + 273 = 205 + 273 = 478^\circ K\\
\dfrac{{{p_1}{V_1}}}{{{T_1}}} = \dfrac{{{p_2}{V_2}}}{{{T_2}}} \Rightarrow {p_2} = \dfrac{{{p_1}{V_1}{T_2}}}{{{V_2}{T_1}}} = \dfrac{{750.{V_1}.478}}{{1,5{V_1}.300}} = 796,6mmHg
\end{array}\)
