Đổi \(\left\{{}\begin{matrix}50^oC=323^oK\\1000cm^3=0,001m^3\end{matrix}\right.\)
Ta có:
\(A=P.\text{∆V}\)
\(40=2.10^5\left(V_2-0,001\right)\)
\(V_2=\dfrac{3}{2500}\)
Áp dụng định luật Gay-luy-xác, ta có:
\(\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}\)
\(\Rightarrow\dfrac{0,001}{T_1}=\dfrac{\dfrac{3}{2500}}{T_1+323}\)
\(\Rightarrow T_1=1615^oK\)