ERT 108/3
PHYSICAL CHEMISTRY
EXERCISES
FIRST LAW OF THERMODYNAMICS
Prepared by:
Pn. Hairul Nazirah Abdul Halim
Question 1
Calculate the work involved in expanding 20.0 L
of an ideal gas to a final volume of 85.0 L against
a constant external pressure of 2.50 bar.
Ans: -16.25 kJ
Solution
To obtain the total work done at constant pressure
when the volume change from Vi to Vf;
w   pex  dV   pex V f  Vi 
Vf
Vi
5
-3
10 Pa
10 m
w  2.50 bar x
x 85.0 - 20.0 L  x
bar
L
w  16.25 kJ
3
Question 2
2.00 mol of an ideal gas undergoes isothermal
reversible expansion from Pi = 25.0 bar and Vi =
4.50 L to Pf = 4.50 bar. Calculate the work of this
process.
Ans: -19.3 x 103 J
Solution
1. Calculate the constant temperature at which the
process is carried out and the final volume.
piVi
25.0 bar x 4.50 L
T

 677 K
-2
-1
-1
nR
2.00 mol x 8.314 x 10 L bar mol K
nRT 2.00 mol x 8.314 x 10 -2 L bar mol -1 K -1 x 677 K
Vf 

 25.0 L
pf
4.50 bar
2. The work of the isothermal reversible expansion;
Vf
w  nRT 
Vi
Vf
dV
 nRT ln
V
Vi
25.0 L
w  2.00 mol x 8.314 J mol K x 677 K x ln
4.50 L
3
w  19.3 x 10 J
-1
-1
Question 3
Relating ΔH and ΔU
The internal energy change when 1.0 mol CaCO3
in the form of calcite converts to aragonite is
+0.21 kJ. Calculate the difference between the
enthalpy change and the change in internal
energy when the pressure is 1.0 bar given that
the densities of the solids are 2.71 g cm-3 and
2.93 g cm-3 , respectively.
Ans: ΔH – ΔU = -0.3 J
Solution
•Calculate the volume of aragonite and calcite.
M

Vm
M
100 g/mol
3
Vm of aragonite 


34
cm
/mol
-3
 2.93 g cm
M
100 g/mol
3
Vm of calcite 

 37cm /mol
-3
 2.71 g cm
• Therefore, 1.0 mol CaCO3 (100 g) as aragonite is 34 cm3,
and that of 1.0 mol CaCO3 as calcite is 37 cm3.
H  U  pV
H  U  pV


H  U  1.0 x 10 Pa x 34 - 37  x 10 m
5
H  U  0.3 Pa.m  0.3 J
3
-6
3
Question 4
When 2.0 mol CO2 is heated at a constant
pressure of 1.25 atm, its temperature increases
from 250 K to 277 K. Given that the molar heat
capacity of CO2(g) at constant pressure is 37.11
J K-1 mol-1. Calculate q, ΔH and ΔU.
Ans: qp= ΔH = 2.0 x 103 J
ΔU = 1.6 x 103 J
Solution
H  q p  C p T  nC p ,m T
H  (2.0 mol) x (37.11 J K -1mol -1 ) x (277 - 250) K
H  q p  2.0 x 103 J
H  U  ( pV )  U  nRT
U  H  nRT
U  (2.0 x 103 J) - (2.0 mol) x (8.314 J K -1mol -1 ) x (277 - 250) K
U  1.6 x 103 J
Question 5
Consider the adiabatic, reversible expansion of
0.20 mol Ar, initially at 250C, from 0.50 L to 1.00 L.
The molar heat capacity of argon at constant
volume is 12.48 J K-1 mol-1. Calculate the work of
adiabatic, wad.
Ans: wad = - 27J
Solution
-1
Cv ,m
-1
12.48 J K mol
c

 1.501
-1
-1
R
8.314 J K mol
 Vi
T f  Ti 
V
 f
1/ c




 0.50 L 
T f  (298 K) x 

1.00
L


1 / 1.501
 188 K
T  188 - 298  - 110 K
w ad  CV T  (0.020 mol) x (12.48 J K -1mol -1 ) x (-110 K)  - 27 J