false

Unlock Document

Chemistry

CHEM 5

Douglas Tobias

Fall

Description

Sabrina Lugo
Section B
11-9-13
Homework #6
#1) The Debye theory of solids provides an accurate description of the temperature
dependance of the heat capacity of monatomic solids over a wide temperature
range. According to the Debye theory, the molar heat capacity at constant volume,
C for an absolute temperature T, is given by the following expres-
v
sion C = 9 R J T N * HQDêTL x *e „x
v QD Ÿ0 He -1L
J
R= gas constant = 8.314
K*mol
QD= the Debye temperature
Given that QD= 309 K for copper, Calc. the molar heat capacity, which will come out
J
in units of K*molfor copper at 103 K using the eq. above.
R = 8.314 J ;
K * mol
Q = 309 K;
T = 103 K;
4 x
T 3 HQêTL x * E
Cv = 9 * R * * ‡ „x;
Q 0 HE - 1L 2
Cv
16.5305 J
K mol
#2) According to the kinetic theory of gases, the distribution of speeds of atoms or
moleucles in a ideal gas is given by the Maxwell-Boltzmann Distribution:
M 3ê2 2 -Mu ë2 RT
f HuL = 4 p I 2 pRT u e
Here, f(u)du is hte probability that an atom/molecule in the gas has a speed between
u and u+du,
M = the molar mass of the atom/molecule
R = the gas constant
T = the absolute temperature
a) Derive an expression for the most-probable speed by solving the equaiton df(u)/d
= 0 for u.
Printed by Wolfram Mathematica Student Edition 2 sclugo_hw6.nb
M 3ê2 2
[email protected]_D := 4 * Pi * * u * E -IM*u MëH2*R*T;
H2 * Pi * R * TL
[email protected];
[email protected];
eq = [email protected]@[email protected], uD ã 0, uD;
[email protected]@3DD
2 R T
:u Ø >
M
b) Calculate an exprssion for the most-probable value of the speed for N 2 (M = 0.028
kg/mol) at T=300K, using
J
R=8.314 so your answer will be in m/s
K*mol
[email protected], T, RD;
value = [email protected]@[email protected], uD ã 0, uD;
0.028 kg 8.314 J
[email protected]@3DD ê. :M Ø , T Ø 300 K, R Ø >
mol K * mol
J
422.087 K K mol
:u Ø >
kg
mol
2
2 K * 2g*m 2
H*Recall J= kg*m \ s *K*mol= m = m*L
s2 Kg s2 s
mol
#3) The equation constant for the reaction:
3Al 2l 6HgL 2Al C3 9HgL
-4
is K p 1.04x 10 at 454 K. A reaction vessel is charged with an initial pressure of
Al Cl of P 0 .If -x is the change in the pressure of Al Cl that occurs as the reac-
2 6 A2 C6 2 6
tion reaches equilibrium, then the law of mass action is:
2 x
J3 N
K p 0 3
IP A2 6lxM
a) Calc. the equilibrium partial pressures of Al 2l 6 and Al3Cl 9 using the usual
0 0
assumption that x is small compared to P A2 C6, using P A2 C6 = 2atm.
Printed by Wolfram Mathematica Student Edition sclugo_hw6.nb 3
Kp = 1.04 * 10 ;
PAl2Cl6 = 2;
2
J H2*xLN
3
equation = : 3 ã Kp>;
HPAl2Cl6L
roots = [email protected], xD
88x Ø -0.0432666;
3
HPAl2Cl6 - xL
root = [email protected], xD
88x Ø -4267.5

More
Less
Related notes for CHEM 5

Join OneClass

Access over 10 million pages of study

documents for 1.3 million courses.

Sign up

Join to view

Continue

Continue
OR

By registering, I agree to the
Terms
and
Privacy Policies

Already have an account?
Log in

Just a few more details

So we can recommend you notes for your school.

Reset Password

Please enter below the email address you registered with and we will send you a link to reset your password.