Ideal gas problem- chemistry.

Question

A flask of fluorine gas is connected to the manometer below ( image of mercury manometer not shown) at 29 degrees C. Calculate the pressure in atm, the molar volume, and the density of the flask if atmospheric pressure os 0.983 atm. 

 

the image on my sheet is a manometer showing a difference in 187 mm where the mercury is lower on the side with the fluorine gas. 

 

Please hep I'm very confused, thanks!!!

Answer 1

187 mm Hg x 1 atm/760 mm Hg = 0.246 atm = change in pressure due to F2 gas

Since the Hg is lower on the side with the gas, this means that the pressure of the gas is > than that of the atmosphere.

Thus pressure of gas = 0.983 atm + 0.246 atm = 1.23 atm 

 

Molar volume = volume of 1 mole F2 gas

PV = nRT 

P = 1.23 atm

V = ? L 

n = 1 mole

R = 0.08206 Latm/Kmole

T = 29ºC = 302 K

Solving for V:

V = nRT/P = (1mol)(0.08206 Latm/Kmol)(302 K)/1.23 atm

V = 20.15 Liters = molar volume

 

Density of gas (not density of flask):

1 mole F2 gas = 20.15 L

1 mole F2 gas = 37.996 g

Density = 37.996 g/20.15 L 

Density = 1.89 g/L (to 3 significant figures)