Question

The pressure of a gas is reduced from 1200.0 mm Hg to 850.0 mm Hg as the volume of its container is increased by moving a piston from 85.0 mL to 350.0 milliliters. What would the final temperature be if the original temperature were 90.0 °C?

Answer 1

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

$\frac{1200×85.0}{363}=\frac{850×350}{T_2}$

$T_2= 1058.75K=785.75°C$