Gas Law Help????

Question

1. A 7.81 x 10^-3 mole sample of oxygen gas was placed in a 0.355 L container at 398 K. What is the pressure exerte by the gas in kPa?

2. A helium balloon with a volume of 410.0 mL is cooled from 48.0 C to -37.0 C. The pressure on the gas is reduced from 110.0 kPa to 91.0 kPa. What is the voulme of the gas at the lower temprature.

Answer 1

1. The ideal gas law is: PV=nRT

Therefore P=nRT/V

n = 7.81x10^-3 mol

R = 8.3145 J/Kmol

T = 398 K

V = .355L * (m^3/1000L) = .000355m^3

P = 7.81x10^-3 mol * (8.3145 J/Kmol) * 398K /.000355m^3 = 72802J/M^3 = 72802 N/m^2 = 72.802 kPa

2. The combined gas law is: P1V1/T1 = P2V2/T2

P1 = 110 kPa

P2 = 91 kPa

T1 = 48C =  (48+273.15)K = 321.15K

T2 = -37C = (-37+273.15)K = 236.15K

V1 = 410 mL

V2 = P1V1T2/(T1P2) = 110*410*236.15/(321.15*91) = 364 mL