httpsblackboarduwindsorcabbcswebdavpid399024dtcontent rid152

https//blackboard.uwindsor.ca/bbcswebdav/pid-399024-dt-content -rid-1529625 12/Ta akehome Tutor ia0%203 pdf Problem #4: Helium gas with a volume V = 260 L under a pressure of 0.180 atm and a temperature of 41 oC is warmed until both pressure and volume are doubled. Determine a) What is the final temperature of the gas? b) Determine how many grams of the helium are there. Use that the molar mass of helium is 4 g/mol and assume that the helium is an ideal gas. Hint: The universal gas constant is 8.314 J/mol-K latm = 101.3kPa, IL-0.00 lm\"

Solution

a) Using the gas equation
P₁V₁/T₁ = P₂V₂/T₂
Here let P₁ = 1.3 atm so P₂ = 2P1 atm;

V₁ = 2.60 L so V₂ = 2V1 L;

T1 = 273 + 41 = 314 K , T2 = ?
Thus T2 = (P₂xV₂ x T₁)/(P₁ x V₁

= (2P₁ x 2V₁ x 314)/(P₁ x V₁

= 1256 K or 983.0 ^oC

b) To find the mass of He use the following equation
PV = nRT where n is the No. of moles of He
i.e. n = PV/RT
P = 1.3 atm, V = 2.60 L, T = 314 K, R = 0.0821 L atm. K^-1 mol^-1
So n = 1.3 x 2.60/0.0821 x 314 = 0.13
Thus the moles of He = 0.13
And the mass of He = 0.13 x 4 = 0.52 g