Saved Bookmarks
| 1. |
When you have learned to integrate derive the Poisson formula for an adiabatic process. |
|
Answer» `m/M C_(mv) dT + pdV = 0`. Differentiating the Mendeleev-Clapeyron equation, we OBTAIN `pdV = V dp = m/M RdT` Eliminating dT from these EXPRESSIONS, we obtain after some transformations `gamma (dV)/V = (dp)/p =0` INTEGRATING , we obtain `gamma int_(v_0)^V (dV)/V + int_(p_0)^(p) (dp)/(p) = 0` From which `gamma ln V - gamma ln V_0+ ln p - ln p_0 = 0 , " or " ln (pV^(gamma)) = ln (p_0v_0^(gamma))` Therefore, `pV^(gamma) = p_0 V_(0)^(gamma)`. |
|