Question

Part A) Find the pressure in mmHg of a 0.124 g sample of helium gas in a 649 mL container at 34 ∘C. Express the pressure to three significant figures and include the appropriate units.

Part B) xture with a total pressure of 770 mmHg contains each of the following gases at the indicated partial pressures: 140 mmHg CO2, 112 mmHg Ar, and 128 mmHg O2. The mixture also contains helium gas.

What is the partial pressure of the helium gas?

What mass of helium gas is present in a 23.0-L sample of this mixture at 263 K ?

Part C)  cylinder with a moveable piston contains 0.76 mol of gas and has a volume of 337 mL .

What is its volume after an additional 0.26 mol of gas is added to the cylinder? (Assume constant temperature and pressure.)

Express your answer to three significant figures.

Part D) A gas mixture with a total pressure of 770 mmHg contains each of the following gases at the indicated partial pressures: 140 mmHg CO2, 112 mmHg Ar, and 128 mmHg O2. The mixture also contains helium gas.

What mass of helium gas is present in a 23.0-L sample of this mixture at 263 K ?

Express your answer in grams.

Answer 1

a) Molecular weight of helium= 4 , moles of helium in 0.124gm, n = 0.124/4=0.031 moles

Volume V= 649ml= 649/1000 =0.649L T= 34 deg.c =34+273.15=307.15K

From PV= nRT

P= nRT/V= 0.031*0.08206(L.atm/mole.K)*307.15/0.649=1.204 atm

b)

Partial pressure of Helium= total pressure- sum of partial pressures of CO2, Ar and O2

= 770-(112+140+128) mm Hg=390 mm Hg

Since partial pressure is the pressure exerted by the gas if it alone occupies the entire volume

V= 23L T= 263K moles of helium , n= PV/RT= (390/760)atm* 23/(0.08206L.atm/mole.K)*263=0.5468

Mass/Molecular weight= 0.5468

Mass of helium= Molecular weight* 0.5468=2.1872 gm

c)

Moles of gas after addition of 0.26ml =0.26+0.76= 1.02mol

from gas law P1V1/n1T1 = P2V2/n2T2

From V1/n1= V2/n2 (since Pressure and temperature remains constant)

337/0.76= V2/1.02

V= 337*1.02/0.76=446.4156ml

d)

Partial pressure of Helium= total pressure- sum of partial pressures of CO2, Ar and O2

= 770-(112+140+128) mm Hg=390 mm Hg

Since partial pressure is the pressure exerted by the gas if it alone occupies the entire volume

V= 23L T= 263K moles of helium , n= PV/RT= (390/760)atm* 23/(0.08206L.atm/mole.K)*263=0.5468

Mass/Molecular weight= 0.5468

Mass of helium= Molecular weight* 0.5468=2.1872 gm