Đáp án:

\(\begin{array}{l}
25.A.500J\\
\Delta U = A + Q =  - 4500 + 5000 = 500J\\
26.C{.23.10^4}J\\
Q = mc\Delta t = 5.460(130 - 30) = {23.10^4}J\\
27.B.4,9kgm/s\\
v = gt = 0,5.9,8 = 4,9m/s\\
\Delta p = m(v - {v_0}) = 1(4,9 - 0) = 4,9kgm/s\\
28.B\\
29.D.N.s\\
\Delta \vec p = \vec F.\Delta t\\
F:N\\
\Delta t:s\\
 \Rightarrow \Delta p:N.s\\
30.B\\
pttt:\dfrac{{pV}}{T}\\
31.A.\dfrac{p}{T} = const\\
32.D\\
\dfrac{{{p_1}{V_1}}}{{{T_1}}} = \dfrac{{{p_2}{V_2}}}{{{T_2}}}\\
 \Rightarrow \dfrac{{{p_2}}}{{{p_1}}} = \dfrac{{{V_1}{T_2}}}{{{V_2}{T_1}}} = \dfrac{{{V_1}.2{T_1}}}{{\dfrac{{{V_1}}}{2}{T_1}}} = 4\\
 \Rightarrow {p_2} = 4{p_1}\\
33.B{.3.10^5}Pa\\
{p_1}{V_1} = {p_2}{V_2} \Rightarrow {2.10^5}.150 = {p_2}.100\\
 \Rightarrow {p_2} = {3.10^5}Pa\\
34.C\\
{p_1}{V_1} = {p_2}{V_2} \Rightarrow \dfrac{{{V_2}}}{{{V_1}}} = \dfrac{{{p_1}}}{{{p_2}}} = \dfrac{4}{{10}} = \dfrac{1}{{2,5}}\\
 \Rightarrow {V_2} = \dfrac{{{V_1}}}{{2,5}}\\
35.C\\
{p_1}{V_1} = {p_2}{V_2} \Rightarrow \dfrac{{{V_2}}}{{{V_1}}} = \dfrac{{{p_1}}}{{{p_2}}} = \dfrac{1}{2}\\
 \Rightarrow {V_2} = \dfrac{{{V_1}}}{2}\\
36.C.\dfrac{{{p_1}}}{{{T_1}}} = \dfrac{{{p_3}}}{{{T_3}}}\\
37.B.546^\circ K\\
{T_1} = 0 + 273 = 273^\circ K\\
\dfrac{{{p_1}{V_1}}}{{{T_1}}} = \dfrac{{{p_2}{V_2}}}{{{T_2}}} \Rightarrow \dfrac{{2.30}}{{273}} = \dfrac{{4.30}}{{{T_2}}}\\
 \Rightarrow {T_2} = 546^\circ K\\
38.A.141,5^\circ K\\
{T_1} = 10 + 273 = 283^\circ K\\
\dfrac{{{V_1}}}{{{T_1}}} = \dfrac{{{V_2}}}{{{T_2}}} \Rightarrow \dfrac{{{V_1}}}{{283}} = \dfrac{{{V_1}}}{{2.{T_2}}}\\
 \Rightarrow {T_2} = 141,5^\circ K\\
39.C.P = Fv\\
P = \dfrac{A}{t} = \dfrac{{Fs}}{t} = Fv\\
40.C\\
41.D\\
42.C.450^\circ K\\
{T_1} = 27 + 273 = 300^\circ K\\
\dfrac{{{p_1}}}{{{T_1}}} = \dfrac{{{p_2}}}{{{T_2}}} \Rightarrow \dfrac{{{p_1}}}{{300}} = \dfrac{{1,5{p_2}}}{{{T_2}}}\\
 \Rightarrow {T_2} = 450^\circ K\\
43.B\\
44.B\\
45.B\\
46.B.227^\circ C\\
{T_1} = 27 + 273 = 300^\circ K\\
\dfrac{{{p_1}}}{{{T_1}}} = \dfrac{{{p_2}}}{{{T_2}}} \Rightarrow \dfrac{{0,6}}{{300}} = \dfrac{1}{{{T_2}}}\\
 \Rightarrow {T_2} = 500^\circ K\\
{t_2} = {T_2} - 273 = 500 - 273 = 227^\circ C\\
47.A.1,{068.10^5}Pa\\
{T_1} = 20 + 273 = 293^\circ K\\
{T_2} = 40 + 273 = 313^\circ K\\
\dfrac{{{p_1}}}{{{T_1}}} = \dfrac{{{p_2}}}{{{T_2}}} \Rightarrow \dfrac{{{{10}^5}}}{{293}} = \dfrac{{{p_2}}}{{313}}\\
 \Rightarrow {p_2} = 1,{068.10^5}Pa\\
48.B.1694,4kJ\\
{Q_1} = \lambda m = 3,{4.10^5}.4 = 1360kJ\\
{Q_2} = mc\Delta t = 4.4180(20 - 0) = 334,4kJ\\
Q = {Q_1} + {Q_2} = 1360 + 334,4 = 1694,4kJ\\
49.D.177^\circ C\\
{T_1} = 27 + 273 = 300^\circ K\\
\dfrac{{{p_1}{V_1}}}{{{T_1}}} = \dfrac{{{p_2}{V_2}}}{{{T_2}}} \Rightarrow \dfrac{{1.2}}{{300}} = \dfrac{{15.0,2}}{{{T_2}}}\\
 \Rightarrow {T_2} = 450^\circ K\\
{t_2} = {T_2} - 273 = 450 - 273 = 177^\circ C\\
50.A\\
51.C.796,6mmHg\\
{T_1} = 27 + 273 = 300^\circ K\\
{T_2} = 205 + 273 = 478^\circ K\\
\dfrac{{{p_1}{V_1}}}{{{T_1}}} = \dfrac{{{p_2}{V_2}}}{{{T_2}}} \Rightarrow \dfrac{{750}}{{300}} = \dfrac{{{p_2}.1,5{V_1}}}{{478}}\\
 \Rightarrow {p_2} = 796,6mmHg\\
52.B\\
53.D
\end{array}\)