Answer / fat cat

I'm guessin 1 bar. Because the pressure will equalise.

Answer / akash varshney

we know that total .no of moles should be constant,
now-
_ _
p1v1=n1RT1 p2v2=n2RT2
now-- after mixing.
_
PV=(n1+n2)RT
from avobe two eqn....
_ _ _
PV=(p1v1/RT1 + p2v2/Rt2)RT
....PV=(p1v1/T1 + p2v2/T2)T
so now for getin the value of final pressure pwe shpuld
know the final value of T .because v=v1+v2.
now p(V1+V2)==(p1v1/T1 + p2v2/T2)T
so p=(p1v1/T1 + p2v2/T2)T/(v1+v2)

if T1=T2=T and V1=V2=V
thn ....
p=(p1v/T+p2v/T)T/(v+v)
so p=(p1+p2)/2
now p=(1.5+.5)/2
p=1 Ans.

Answer / sunil

pressure changes still they come to thermal eqilibrium .
final volume=v1+v2 use p*v=n*R*t=====>T=PV/NR
FINAL TEMPERATURE T=P(V1+V2)/NR
P(V1+V2)/NR= (P1V1+P2V2)/NR
FINAL PRESSURE=(P1V1+P2V2)/(V1+V2)
SO TO CONCLUDE EITHER VOLUME OR TEMPERATURE OF
VESSELES MUST BE GIVEN.
IF VOLUME OF BOTH THE VESSELS IS SAME
FINAL PRESSURE=(P1+P2)V/2V
====>P=(P1+P2)/2=1bar

Answer / arun.k

pressure will become equal in both vessels.

Answer / sandeep

it will depend on temperature and volume of both the
cylinders..so we cant guess answer without these data.

Answer / wqweqweqw

above all answers are not correct

Answer / lokwadermmm jntucep

hallow plz give d volumes of cylinders

by taking same volume
p1v1=p2v2
den easyyyyyyyyyyyyyyyy
p1=p2=1

Answer / vivek

1 BAR coz after connect net
pressure in opposite will be
equal so no movement of gas
will be there after reachng 1
bar..this is equilibrium state.

Answer / rakesh swami

Assuming the initial temperatures to be different and the cylinders (and valve, piping too) to be insulated, we can put entropy condition,i.e., overall entropy is zero for reversible adiabatic process, and get final temperature. Now all we need to do is to get final pressure by using ideal gas law PV=mRT, m=m1+m2, V=V1+V2.

Answer / abel

It will depend on the relative volumes of the cylinders.

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