- #1

- 94

- 0

A flask with a volume of 1.50 L, provided with a stopcock, contains ethane gas (C2H6) at 300 K and atmospheric pressure (1.013×10^5 Pa). The molar mass of ethane is 30.1 g/mol. The system is warmed to a temperature of 380 K, with the stopcock open to the atmosphere. The stopcock is then closed, and the flask cooled to its original temperature. a) What is the final pressure of the ethane in the flask? b)How many grams of ethane remain in the flask?

------------------

V=.0015m^3

T1=300k

T2=380k

V1=1.013x10^5 PA

So, using PV=nRT I solve for n getting .0609 mol, this leads me to believe that the volume did not change much(right?)

So

V1P1/T1=V2P2/T2

Assuming V1=V2, P1/T1=P2/T2

Solving for P2 i get 128313 Pa or 1.28x10^5 Pa,

I am lost

Can anyone please help?

I will be in your debt forever :)