What will be the pressure of a sample of 48.0 grams of oxygen gas in a glass container of volume 5.2 L at 25 degrees Celsius?

Answer 1

Molar Mass O2: 32.00, Molar Mass CO2: 44.01, T= 21C+273= 294Kelvins V=5.00 L Grams of O2=20.0 so moles O2= 20.0g/32.00g=.625g/mol

Grams of CO2=4.4 so moles CO2= 4.4g/ 44.01=.1

Total moles: .625+.1= .725 mols (this is n)

Ideal Gas Law: PV=nRT....Rearrange and solve for pressure.

P=nRT/V.....R is the constant which is equal to .08206L*atm*mol

P=(.725)(.08206)(294K)/(5.00L)

P= 3.5atm

Answer 2

PV = nRTP = pressure = ?

V = volume = 5.2 L

n = moles of gas = 48 g x 1 mole/32 g = 1.5 moles O2 gas

R = gas constant = 0.0821 L-atm/K-mole

T = temperature in K = 25 + 273 = 298 K

Solving for P = nRT/V = (1.5)(0.0821)(298)/5.2

P = 7.0 atm (to 2 significant figures)

Answer 3

The pressure is 0,14 atmosphere.

Answer 4

3.87 ATM

Answer 5

7.88 atm