LEMNISCATA
Matemàtiques, física, química…
\[\bar{C}_{p(O_2)} = 6,095 + 3,253 \cdot 10^{-3} T – 1,017 \cdot 10^{-6} T^2 \ (\mathrm{cal/K \ mol})\]Com que \( Q_p = \Delta H \), i si expressem \( H = f(T, P) \), per ser \( dP = 0 \Rightarrow dH = \left( \frac{\partial H}{\partial T} \right)_P dT + \left( \frac{\partial H}{\partial P} \right)_T dP = C_p dT \)\[\int dH = \int n \bar{C}_p dT = n \int (6,095 + 3,253 \cdot 10^{-3} T – 1,017 \cdot 10^{-6} T^2) dT\]\[\Delta H = 1 \ \mathrm{mol} \left[ 6,095 (T_2 – T_1) + \frac{3,253 \cdot 10^{-3}}{2} (T_2^2 – T_1^2) – \frac{1,017 \cdot 10^{-6}}{3} (T_2^3 – T_1^3) \right] = 633,9 \ \mathrm{cal}\]