INSTRUCTOR: Here I'm going to
apply the first law of thermodynamics
for a closed system, ideal gas,
in which we have an adiabatic reversible process,
so I tried to represent this here on the left.
For example, if we're carrying out compression,
we start with a small weight,
and as we compress the gas, we keep adding more weight,
so you'll notice the pressure of the gas
in this case, 1.26 megapascals,
is the same as the external pressure being exerted on it
because it's reversed well we could also
just as easily be looking at expansion
where we start at 1.26 megapascals
and then expand the gas reversibly
by removing weights till we end up at 0.26 megapascals.
If we write down the first law for this system
in differential form,
the change in internal energy
change in expansion compression work
Q is zero because it's adiabatic
and we have no shaft work in this system.
Because it's an ideal gas, dU is Cv dT.
Because it's reversible,
I can write this differential as P dV
where P is the pressure of the gas
it's also the external pressure,
and then I can make a substitution using the ideal gas law,
and I can replace pressure as RT over the volume dV.
The left side's not changed,
so I can rearrange this, have the
temperature terms on the left side,
and volume terms on the right,
and then I can integrate from T1 to T2
from volume one to volume two.
Cv is a constant, so we're looking the case Cv is constant.
It's a restriction on this system.
Cv then we have log of T2 over T1.
On the right side, R is a constant of course,
and we have the log of V2 over V1.
Now I'm gonna use the ideal gas law
on the right side to rewrite V2 over V1
as T2 over T1 P1 over P2.
So all I've done is take the
ideal gas law P2 V2 equals R T2,
and let's replace V2 by R T2 over P2.
Likewise, for V1, the Rs cancel.
I end up with this log term on the right side.
But notice I rewrite that and have a log of products
I can write this as the sum of logs,
so, and important I don't leave out this minus sign,
so I have minus R log T2 over T1
minus R log P1 over P2,
so I'm gonna bring this term to the left side
and factor out the log term,
so the left side becomes Cv plus R
and then the log of T2 over T1,
and on the right side,
I'm gonna rewrite this term by reversing P1 and P2
and changing the minus sign to a plus sign,
so again, just a property of logs.
The right side becomes plus sign now log of P2 over P1.
Well, Cv plus R for an ideal gas is Cp.
What I'm gonna do is divide both sides by Cp,
so this is one, and, again, using property of logs,
R over Cp becomes an exponent.
Now if I take an exponent of both side of this equation,
I have T2 over T1 equals P2 over P1
raised to the power of R over Cp,
and so this is the equation that we're interested in.
Important to remember it's only
reversible, adiabatic, constant Cp,
and we derived it for a closed system.
It turns out it's also gonna be true
for flow system that has the same attributes.
And so, as we pressurize the gas,
this tells if we do it adiabatically,
the temperature's gonna increase
if we expand it adiabatically,
the temperature's gonna decrease.