# If the initial temperature of an ideal gas at 2.250 atm is 62.00 °C, what final temperature would cause the pressure to be reduced to 1.600 atm?

Nov 23, 2017

$\text{238.2 Kelvin}$

#### Explanation:

Gay-Lussac's law: For a given mass and constant volume of an ideal gas, the pressure exerted on the sides of its container is directly proportional to its absolute temperature.

Mathematically,

P ∝ T

${P}_{f} / {P}_{i} = {T}_{f} / {T}_{i}$

T_f = T_i × P_f/P_i

${T}_{f} = \text{(62.00 + 273.15) Kelvin" × (1.600 cancel"atm")/(2.250 cancel"atm") = "238.2 Kelvin}$