Question

A sample of oxygen gas initially at 0.760 atm is cooled from 27°C to −93°C at constant volume.

What is its final pressure (in atm)?

Answer 1

0.38 atm

Step-by-step explanation:

Gay-Lussac's Law states:

`(P_1)/(T_1) =(P_2)/(T_2)`

p = pressure

T = temperature

Change temps from Celcus to Kelvin:

27+273 = 360

-93+273 = 180

plugging the information in we have:

`(0.760 atm)/(360K) = (x)/(180K)`

Cross multiply:

136.8 = 360x

divide 136.8 by 360 = .38 atm