Question

what is the volume of a gas at 2.00 atm and 200.0 K if its original volume was 300.0 L at 0.250 atm and 400.0 K

Answer 1

Given:
P1 = 2.00 atm
T1 = 200.0 K
V2 = 300.0 L
P2 = 0.250 atm
T2 = 400.0 K

Required:
V1

Solution:
P1V1/T1 = P2V2/T2
(2.00 atm) V1 / (200.0 K) = (300.0 L) (0.250 atm) / (400.0 K)
V1 = 18.75 L