Question

22. An open flask has Helium gas at 2atm and
327°C. The flask is heated to 527°C at the
same pressure. Th fraction of original gas
remaininig in the flask is
1) 3/4
2) 1/4 3) 1/2
4) 2/5​

Answer 1

Explanation:

from problem

P1=1atm,TE=327℃=600K

PE=1atm,TE=527℃=800k

Formula P1V1/TE=P2V2/TE

from above Eqn V2=P1×V1×T2/T1×P2

V2=1×V1×800/600×1

V2= 4/3V1

Volume of air expelled=V2-V1=4/3V1-V1=V1/3

Fraction of air expelled out= V1/3/4V1/3=1/4

Fraction of air remaining in the flask =1-1/4=3/4.