[tex]Dane:\\t_1 = 27^{o}C\\T_1 = (27+273) K = 300 \ K\\V_2 = 3V_1\\Szukane:\\T_2 = ?[/tex]
Z prawa Gay-Lussaca:
[tex]\frac{V_1}{T_1} = \frac{V_2}{T_2}\\\\ale\\\\V_2 = 3V_1, \ zatem:\\\\\frac{V_1}{T_1} = \frac{3V_1}{T_2} \ \ \ |:V_1\\\\\frac{1}{T_1} = \frac{3}{T_2}\\\\T_2 = 3T_1\\\\T_2 = 3\cdot300 \ K = 900 \ K\\\\900 \ K = (900-273)^{o}C = 627^{0}C\\\\\boxed{T_2 = 900 \ K = 627^{o}C}[/tex]
Odp. Temperatura końcowa powietrza wynosi 627°C.