This is the expression valid for van der Waals gases. Calculate the final temperature if the Joule’s free expansion experiment was carried out to double the volume of 1 L gas at 25 °C with a) He b) CO2 molecules. The van der Waals coefficients, a, for He and CO2 are 0.0346 and 3.658, respectively. (Hint: Use equipartition to calculate Cv values and assume three of the vibrations not populated in carbon dioxide)
This is the expression valid for van der Waals gases. Calculate the final temperature if the Joule’s free expansion experiment was carried out to double the volume of 1 L gas at 25 °C with a) He b) CO2 molecules. The van der Waals coefficients, a, for He and CO2 are 0.0346 and 3.658, respectively. (Hint: Use equipartition to calculate Cv values and assume three of the vibrations not populated in carbon dioxide)
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This is the expression valid for van der Waals gases. Calculate the final temperature if the Joule’s free expansion experiment was carried out to double the volume of 1 L gas at 25 °C with a) He b) CO2 molecules. The van der Waals coefficients, a, for He and CO2 are 0.0346 and 3.658, respectively. (Hint: Use equipartition to calculate Cv values and assume three of the vibrations not populated in carbon dioxide)
Transcribed Image Text:dU = C,dT
In this question, your task will be to correct Joule and find the correct value for the Joule
coefficient for a gas that obeys van der Waals equation.
Let's bring the entropy in to the discussion. As we know from either statistical or
thermodynamics definition of entropy if any gas is expanded the entropy increases.
as - (),v +(), a.
dV +
dT.
Let us look at these three terms in turn. First term is ds;
Since the internal energy is constant in Joule experiment:
Tds – PaV -0.
PaV
ds =
as
Second term is
We make use of the Maxwell relation:
(),- (),
as
as
The third term is
(),- ()G),- (H)/(), -4
as
=
Now let's write the initial equation with these new quantities:
PdV
CydT
aV +
T
The equation can be reorganized as:
P =T
+ Cy
from which we immediately obtain:
ƏT
1
Р-Т
Cy
de
av
This is the quantity we measure with the Joule experiment. When we use the van der Waals
equation in the pressure terms of the equation above, we get:
-a
Gy. V²
(av),
= -Cu
= -Cy
dU = CydT +
This equation can be integrated over temperature and volume change:
dU =
CydT +
= 0
Assuming Cvis independent of temperature, the equation becomes:
Cy (T2 – T,) = a
AT
Transcribed Image Text:6. The first law of the thermodynamics state that J-daW-CTh. This equation
implies that the internal energy is a fùnction of T and V. Using the mathematical definition
of a function with two variables:
du -
dT
AP
In this expression, two partial derivatives must be determined (measured): a- how the internal
energy changes with temperature at constant volume and b- how the internal energy changes with
volume at constant temperature. In the first case (constant V):
w = 0 = dU = đạy
Constant V
ne,
du =
dT-
AP
đg, =
dT
đgy = CydT =
= Cy
heat capacity of any gas-at constant volume can be measured using calorimeters. So the energy
equation becomes:
ne,
dU = C,dT +
dV
Joule has attempted to determine the internal energy change with volume at constant
temperature and set up following experiment.
gas
vac
gas (pi, Ti, И) - дas (pг, Tz, V)
He put the entire system in an isolated system, so that no heat transfer occurs between system
and surroundings (q-0). He also created a vacuum as shown in figure, so that when the system
freely expands (Pa) without doing any work (W=0). Therefore, the internal energy change for
such experiment must be dU-0 according to the first law. Thus:
Since g= w = 0
dU or AU= 0
Constant U
Recall
du = C,dT +
dv
measure in Joule exp't!
>-4r>-ק
'AT
Joule did this,
= 12;
or
Joule coefficient
His experimental setup allowed him to measure temperature change upon expansion, which
must have occurred at constant internal energy. The gas he used was a nearly perfect gas and the
temperature change was too small to observe in his scale. Therefore, he made a mistake and
concluded that no temperature change occurred during such an expansion. However, his
conclusion is valid and true for ideal gases. Today, sometimes we use his conclusion to define the
first law of the thermodynamics "The internal energy of an ideal gas depends only
temperature"
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