Đáp án:
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Giải thích các bước giải:
câu 20:
\({T_2} = 293K;{P_2} ={ P_1}\frac{1}{{40}}{P_1};\)
đẳng tích:
\(\frac{{{P_1}}}{{{T_1}}} = \frac{{{P_2}}}{{{T_2}}} \Rightarrow {T_2} = {T_1}({P_1} + \frac{{{P_1}}}{{40}})\frac{1}{{{P_1}}} = 293.(1 + \frac{1}{{40}}) = 300K \Rightarrow {t_1} = {27^0}C\)
bài 21:
\({t_1} = {15^0}C;{T_1} = 288K;{T_2} = 573K;\)
áp suất tăng
\(frac{{{P_1}}}{{{T_1}}} = \frac{{{P_2}}}{{{T_2}}} \Rightarrow \frac{{{P_2}}}{{{P_1}}} = \frac{{{T_2}}}{{{T_1}}} = \frac{{573}}{{288}} \approx 2\)
bài 22:
\({T_1} = 305K;{T_2} = 390K;{V_2} = {V_1} + 1,7l\)
đẳng áp:
\(\frac{{{V_1}}}{{{T_1}}} = \frac{{{V_2}}}{{{T_2}}} \Leftrightarrow \frac{{{V_1}}}{{305}} = \frac{{{V_1} + 1,7}}{{390}} \Rightarrow \left\{ \begin{array}{l}
{V_1} = 6,1l\\
{V_2} = 7,8l
\end{array} \right.\)