Download Exam 1 Solutions – 100 points

Chemistry 360
Fall 2014
Dr. Jean M. Standard
September 17, 2014
Name _________KEY__________________
Exam 1 Solutions – 100 points
R = 0.08206 L atm mol–1 K–1 = 8.314 J mol–1 K–1 = 0.08314 L bar mol–1 K–1
1 atm = 101325 Pa = 1.013 bar = 760 torr; 1 bar = 105 Pa = 0.98692 atm
1 L atm = 101.3 J; 1 L bar = 100 J
1.)
(14 points) Sketch a plot illustrating the typical behavior of the compression factor Z as a function of P
for a real gas. Make sure that you indicate the ideal gas value. Briefly explain the physical reasons for
the regions of positive and negative deviations from ideality.
A typical plot of Z vs. P for a real gas at a particular is shown below. The ideal gas result, Z=1, is indicated on
the graph.
Z
Z=1
P
Note that the scale of P on the x-axis usually ranges from P=0 to P=1000 atm or so.
The region of negative deviation for a real gas typically occurs at low pressure. This is a result of
intermolecular interactions. At low pressure, the atoms or molecules of a real gas interact with one another and
"stick together". This results in a pressure that is lower than expected, and therefore a compressibility factor Z
that is less than 1.
The region of positive deviation for a real gas occurs at high pressure. This is a result of the finite size of real
gases. An ideal gas takes up no space, but for a real gas, as the pressure increases, the atoms or molecules get
crushed together and cannot get any closer. This leads to a volume that is larger than for the ideal gas case, and
therefore a compressibility factor Z that is greater than 1.
2
2.) (14 points) Determine the quantities q, w, and ΔU for the following processes involving one mole of an
ideal monatomic gas (helium). Express your results in Joules.
a.) One mole of helium gas at 200ºC and 2.0 bar is expanded reversibly and isothermally to a final
pressure of 0.25 bar.
Starting from the basic definition of work, we have
w = −
∫ Pext dV
This process is reversible. Therefore, Pext = P . This yields
€
w = −
€
Using the ideal gas equation, we have P =
€
V2
P dV .
V1
nRT
. Substituting,
V
w = −
€
∫
∫
V2
V1
nRT
dV .
V
Since the process is isothermal, we can pull out T along with the other constants and integrate the
remaining expression.
€
w = − nRT
∫
V2
V1
1
dV
V
$V '
w = − nRT ln& 2 ) .
% V1 (
We need to determine the initial and final volumes. The ideal gas equation can be utilized for this purpose.
The initial volume is
€
V1 =
nRT1
P1
(1mol) (0.08314 L bar mol−1K −1 ) ( 473.15K )
=
(2.0 bar )
V1 = 19.7L.
Noting that the initial and final temperatures are the same, the final volume is
V2 =
=
nRT2
P2
(1mol) (0.08314 L bar mol−1K −1 ) ( 473.15K )
(0.25bar )
V2 = 157.4 L.
3
2 a.)
Continued
The work in Joules can then be determined by substitution,
"V %
w = − nRT ln $ 2 '
# V1 &
" 157.4 L %
= − (1mol) 8.314 J mol−1K −1 ( 473.15K ) ln $
'
# 19.7L &
w = − 8175J.
(
)
To determine the change in internal energy, we know that for an ideal gas,
dU = C v dT ,
or in integrated form (given that C v is independent of temperature),
€
ΔU = C v ΔT .
€
Since the process is isothermal, ΔT = 0 , and therefore we know that
€
ΔU = 0 .
€
The heat q may then be obtained using the First Law,
€
ΔU = q + w .
With ΔU = 0 , we have that
€
q = − w = 8175 J.
€
2 b.)
One mole of helium gas at 200ºC and 2.0 bar is expanded isothermally against a constant external
pressure of 0.25 bar.
Again, starting from the basic definition of work, we have
w = −
∫ Pext dV .
In this case, the process is not reversible. However, the process involves an expansion against constant
external pressure, so we can pull the factor of Pext out of the integral. This yields
€
w = − Pext
€
∫
V2
V1
dV .
Integrating leads to the expression
€
€
w = − Pext (V2 − V1) .
4
2 b.)
Continued
We again need to determine the initial and final volumes. Since the initial number of moles, pressure, and
temperature are the same as in part (a), the initial volume is calculated from the ideal gas equation in the
same way as before,
V1 =
nRT1
P1
(1mol) (0.08314 L bar mol−1K −1 ) ( 473.15K )
=
(2.0 bar )
V1 = 19.7L.
Since the expansion will occur against the constant external pressure until the point that the final pressure
P2 is equal to Pext , the final volume can be calculated as before,
V2 =
€
€
nRT2
P2
(1mol) (0.08314 L bar mol−1K −1 ) ( 473.15K )
=
(0.25bar )
V2 = 157.4 L.
Finally, the work in Joules can be determined,
w = − Pext (V2 − V1 )
= − ( 0.25bar ) (157.4 L − 19.7L )
" 100 J %
= − 34.43L bar $
'
# 1L bar &
w = − 3443J.
To determine the change in internal energy, we know that for an ideal gas,
dU = C v dT ,
or in integrated form (given that C v is independent of temperature),
€
ΔU = C v ΔT .
€
Since the process is isothermal, ΔT = 0 , and therefore we know that
€
ΔU = 0 .
€
The heat q may then be obtained using the First Law,
€
ΔU = q + w .
With ΔU = 0 , we have that
€
€
q = − w = 3443 J.
5
3.) (14 points) The entropy S is a state function that plays a key role in the Second and Third Laws of
Thermodynamics.
a.) A proposed differential form for the state function S in terms of pressure and temperature is given
by
#
3&
dS = 2PT 2 dP + % 2P 2T −
( dT .
$
T'
Determine whether or not the differential is exact.
€
2
3
For this problem, we have A( P,T ) = 2PT and B( P,T ) = 2P 2T −
. The test for exactness is that
T
#∂ A &
#∂ B &
%
( = %
( .
$ ∂ P 'T
€ $ ∂ T 'P
€
Evaluating these partial derivatives yields
# ∂€A &
%
( = 4PT
$ ∂ T 'P
and
#∂ B &
%
( = 4PT .
$ ∂ P 'T
Since these quantities are equal, the differential is EXACT.
€
€
b.) A researcher in thermodynamics proposed the following form for the temperature and volume
dependence of S for a real gas,
S(T ,V ) = nRT lnV + nRT 2 .
Construct an exact differential for S based on this function.
€
An exact differential of the function f ( x, y ) has the form
#∂ f &
#∂ f &
df = %
( dx + %
( dy .
$ ∂ x 'y
$ ∂ y 'x
€
Thus, for a function S(T ,V ) , the exact differential is given by
€
#∂ S &
#∂ S &
dS = %
( dT + %
( dV .
$ ∂ T 'V
$ ∂ V 'T
€
Evaluating these partial derivatives for the function given above yields
# ∂€S &
%
( = nR lnV + 2nRT
$ ∂ T 'V
and
#∂ S &
nRT
.
%
( =
V
$ ∂ V 'T
Substituting, the exact differential is
€
€
" nRT %
dS = ( nR lnV + 2nRT ) dT + $
' dV .
# V &
€
6
4.) (15 points) True/false, short answer, multiple choice.
a.) True or False : An open system allows matter but not energy to be exchanged between the system and
surroundings.
b.) True or False : The Zeroth Law of thermodynamics is a statement about chemical equilibrium in
molecular systems.
c.) Short answer
The ____critical point________ is the location on a pressure-volume diagram where condensation first
occurs as the system is cooled.
d.) Short answer
A function that possesses an exact differential is one that is said to be path _____independent_________ .
e.) Multiple Choice: The isotherms of a real gas are expected to be very similar to those of an ideal gas under
what conditions?
1) low molar volume.
2) low temperature.
3) high pressure.
4) high temperature.
7
5.) (14 points) One mole of methane gas is placed into a bulb with a volume of 600 mL at 100°C.
a.) Determine the pressure in atm exerted by the gas assuming that methane behaves as an ideal gas.
Assuming that the gas obeys the ideal gas equation, we have
P=
=
nRT
V
(1mol) (0.08206 L atm mol−1K −1 ) (373.15K )
(0.600 L )
P = 51.0 atm.
b.) Determine the pressure in atm exerted by the gas assuming that methane behaves according to the
van der Waals equation. The van der Waals constants for methane are a = 2.253 L2 atm mol–2 and
b = 0.0428 L mol–1.
Treating the gas as obeying the van der Waals equation, we have
P=
=
nRT
an 2
− 2
V − nb V
(1mol) (0.08206 L atm mol−1K −1 ) (373.15K )
0.600 L − (1mol) ( 0.0428L mol−1 )
(2.253L atm mol ) (1mol)
2
−
−2
(0.600 L )
2
2
P = 48.7atm.
c.) What is the primary physical reason for the difference in pressure observed for the van der Waals
gas and the ideal gas in parts (a) and (b)?
We see that the pressure calculated for the van der Waals gas is lower than the pressure calculated for the
ideal gas. As a result, the compressibility factor Z for the van der Waals gas would be less than 1. This is
the region where attractive intermolecular interactions lead to deviations from ideality.
8
6.) (14 points) A sample of 10.0 moles of liquid water at 300 K is mixed with a sample of 20.0 moles of liquid
water at 360 K in an insulated container at 1 bar constant pressure.
a.) Determine the final temperature of the water, assuming that it may be determined as a molefraction-weighted average of the temperatures of the two samples.
The total number of moles once the samples are mixed is 30.0 moles. The mole fraction of the first sample
( x1 ) is
x1 = 10.0 mol 1
= .
30.0 mol 3
The mole fraction of the second sample ( x2 ) is
x2 = 20.0 mol 2
= .
30.0 mol 3
The final temperature of the water is therefore
T f = x1T1 + x2T
1
2
(300 K ) + (360 K )
3
3
T f = 340 K.
= b.) Calculate the change in enthalpy for the mixing process in Joules. It is easiest to break the system up
into two parts (the 10.0 mole sample and the 20.0 mole sample) and calculate the enthalpy changes of
the two parts separately. The overall enthalpy change is then just the sum of the two parts. Assume
that the molar constant pressure heat capacity of liquid H2O is independent of temperature and
given by C p,m = 75.3 J mol−1K −1 .
Since the system is at constant pressure, the heat evolved or absorbed in either part is equal to the enthalpy
change,
q p = ΔH.
The exact differential of H is given by
" ∂H %
dH = C p dT + $
' dP .
# ∂P &T
At constant pressure, the second term drops out (since dP=0) to give
dH = C p dT.
For constant heat capacity, integrating both sides yields
ΔH = C p ΔT.
9
6 b.)
Continued
The overall enthalpy change can be broken up into the heat absorbed by the cold water (300 K, sample 1)
and the heat given off by the hot water (360 K, sample 2),
ΔH = ΔH1 + ΔH 2 .
Calculating the enthalpy change for sample 1, we have
ΔH1 = C p (T f − T1 ) = nC p,m (T f − T1 )
(
)
= (10.0 mol) 75.3 Jmol–1K –1 (340 K − 300 K )
ΔH1 = 30120 J .
Calculating the enthalpy change for sample 2, we have
ΔH 2 = C p (T f − T2 ) = nC p,m (T f − T2 )
(
)
= ( 20.0 mol) 75.3 Jmol–1K –1 (340 K − 360 K )
ΔH 2 = –30120 J .
Finally, computing the overall enthalpy change gives
ΔH = ΔH1 + ΔH 2
= 30120 J + (−30120 J ) ΔH = 0 .
This makes sense because the overall q for the process equals 0 (it occurs in an insulated container), and the
process occurs at constant pressure. Therefore,
ΔH = q p = 0.
€
10
7.) (15 points) True/false, short answer, multiple choice.
a.) True or False: The work done by stretching a rubber band is given by dw = FdL , where F is the tension
of the rubber band and L is the length of the rubber band.
€
b.) True or False : The enthalpy H is a state function that corresponds to the heat evolved at constant
volume, qV.
c.) Short answer
An __adiabatic____________ process is one in which there is no flow of heat between system and
surroundings.
d.) Short answer
The ____Equipartition Theorem______ states that each degree of freedom of a gas contributes 12 nRT to
the internal energy U of the system.
€
e.) Multiple Choice: Circle the correct result for a reversible isobaric process in which an ideal gas undergoes
an expansion from 1 L to 10 L.
1) w > 0 , ΔU > 0 , q > 0.
2) w > 0 , ΔU < 0 , q < 0.
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3) w < 0 , ΔU > 0 , q > 0.
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4) w < 0 , ΔU < 0 , q < 0.
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