Adiabatic Process - Thermodynamics Video Lecture - Class 11

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watched all the videos from part 1 to 11
before going ahead with part 12 so as I
already told adiabatic process is the
one in which there is no heat flow
between the system and the surrounding
so in case of an ideal gas we looked at
the isothermal process for an ideal gas
now let us look at the adiabatic process
the equation for an ideal gas is PV to
the power gamma is equal to constant now
what is gamma here gamma is nothing but
specific heat ratio you remember I spoke
about gamma in some of the previous
slides where gamma is nothing but CP by
CV right so this gamma is nothing but
the specific heat ratio so the relation
between there in case of isothermal
process the relation was PV is constant
here pv to the power gamma is constant
now you do not bother about how we reach
this equation that you will study in
your higher classes but for now you can
just remember this as I told gamma is
the specific heat ratio now similar to
isothermal curves we will have adiabatic
curves so adiabatic curve is also
nothing but a plot between pressure and
volume so even here pressure and volume
are inversely proportional to each other
so that curve appears somewhat like this
so the curve does look wise the curves
for isothermal as well as adiabatic
processes look similar but there is a
difference we will look at the
difference as we go ahead and solve
numerical problems now let us look at
adiabatic change of an ideal gas as I
told during isothermal change isothermal
expansion or contraction there was some
work which was involved similarly during
adiabatic change of an ideal gas also
there is some work done which is
involved so work done during adiabatic
change of an ID
gasps here also let us consider an ideal
gas witch whose initial state is
determined by p1 v1 t1 because here in
this case none of these three is our
constant and let us see its final state
is defined by p2 v2 and t2 so for this
gas to reach from this initial
temperature to this final temperature
there has to be some work done now
similar to isothermal process here also
we can say that work n is nothing but P
into DV where V is nothing but changing
from v1 to v2 right so we can write this
as P integration of DV from v1 to v2 now
the equation of State for adiabatic
change is PV to the power gamma is equal
to constant so we can see that P is
equal to some constant divided by V to
the power gamma so now we can put this
value of P in the above equation so we
get instead of P we can write constant
divided by V to the power gamma so this
V to the power gamma will go inside so
this will be integrated from v1 to v2 so
we can write it as some constant now
what will be the integration of V to the
power minus gamma it would be V to the
power minus gamma plus 1 divided by
minus gamma plus 1 from v1 to v2 so this
we can write as some constant divided by
1 minus gamma so inside we have V 2 to
the power 1 minus gamma minus V 1 to the
power 1 minus gamma so this
can write as some constant divided by
one minus gamma 1 by V to the power
gamma minus 1 minus 1 by V 1 to the
power gamma minus 1 now we know that
this constant is nothing but PV to the
power gamma so we can say that P 1 V 1
to the power gamma is equal to P 2 V 2
to the power gamma right so the same
expression can be written in this way we
can write this expression as this one
instead of this constant I can write P 2
V 2 to the power gamma divided by V 2 to
the power gamma minus 1 minus P 1 V 1 to
the power gamma divided by V 1 to the
power gamma minus 1 inside bracket and I
keep 1 minus gamma outside so what I did
this constant I entered here as well as
here and instead of the constant I wrote
V 2 V 2 to the power gamma and P 1 V 1
to the power gamma so this we can write
as this expression can be written as 1
by gamma minus 1 and inside it will be P
2 V 2 minus T 1 V 1 now we also know
that P V can be written as RT for an
ideal gas for any ideal gas the equation
of state will remain the same so we can
write P 2 V 2 is equal to RT 2 and P 1 V
1 as at T 1 so we get this expression so
that means the final expression which we
get for work done is equal to 1 by gamma
minus 1 ad I take outside T 2 minus T 1
so this is the work done during
adiabatic change of an ideal gas because
during adiabatic change both volume
pressure as well as temperature
change from one value to another as we
grew from initial to final state so this
is the work that for an adiabatic change
where t1 is the initial temperature and
t2 is the final temperature now even in
this case let us to look at the two
different scenarios the first scenario
is when work done is positive well then
will be positive when t1 is greater than
t2 because when t1 is greater than p2
this quantity inside the bracket will be
positive therefore work done will be
positive that is work done by the gas
when initial temperature is greater than
final temperature so therefore in this
case temperature of the gas decreases in
the next case where work done is less
than 0 it will be less than 0 only when
t1 is less than t2 that means initial
temperature should be less than final
temperature so here in this case the
temperature of the gas increases so when
the temperature of the gas decreases in
that case work is done by the gas when
the temperature of the gas increases
then work is done on the gas so that was
all about adiabatic thank you please
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