Tutorial 1
By
Miss Anis Atikah Ahmad
Question 1
Calculate CP of 586 g of CH4 (g) at 2000K and
1 bar if CP,m of CH4 at 2000 K and 1 bar is 94.4
Jmol-1K-1
Question 1
• Recall
C P ,m 
Cp
n
Value is not
given
1
CP,m  94.4 J  mol  K
1
• Find n;
mass
586 g
n

 36.625mol
1
MW 16 g  mol
Thus;
C p  36.635mol 94.4 J  mol  K  3457.4 J  K
 3.46kJ  K 1
1
1
1
Question 2
Adam and Anthony found that, U P   6.08Jmol
for air at 28°C and pressures in the range 1 to
40 atm. Calculate U m Vm T for air at:
a) 28°C and 1.00 atm
b) 28°C and 2.00 atm
m
T
1
atm 1
Question 2
Given;
 U m 
 U m 


  6.08J / mol , Find
 P T
 Vm T
 U m 

 can be written as;
 Vm T
 U m   U m   P 

  
 

 Vm T  P T  V T
Assuming air as ideal gas,
PV  nRT
PV n  RT
PVm  RT
P  RT Vm
Thus;
 P 

   RT Vm2
 Vm T
Value is not
given
P 2Vm2  R 2T 2
Vm2  R 2T 2 P 2
 P 

   RT R 2T 2 P 2
 Vm T
 P 

   P 2 RT
 Vm T


 U m   U m 
2

 
  P RT
 V T  P T


Question 2
 U m   U m 
2

 
  P RT
 V T  P T


(a) Extract out the data & substitute in the equation:
P  1atm
T  28C  28  273.15  301.15K
R  8.314 J K  mol
 U m 
2
1
1

  6.08 J mol  atm 1atm 8.315JK mol 301.15K 
 V T
 2.42 10 3 atm

(b) Doubling P;

 U m 
3
3

  2.42 10 atm  4  9.7110 atm
 V T

Question 3
One mole of He gas essentially independent of temperature
expands reversibly from 24.6 L and 300 K to 49.2 L. Calculate
the final pressure and the temperature if the expansion is:
(a) isothermal
(b) adiabatic
Sketch these two processes on a P-V diagram
Question 3
• Extract out the information:
V1  24.6L
n  1mol
V2  49.2L
CV ,m  3R 2
(a) Isothermal process; ΔT=0
Thus; T1  T2  300K
PV  nRT
P2  nRT2 V2


 1mol 8.206 102 L  atm K  mol 300 K  49.2 L
 0.500atm
Question 3
(b) Adiabatic process; q=0
For reversible adiabatic process;
P1V1  P2V2 where   C P ,m CV ,m
  CP ,m CV ,m  CV ,m  R  CV ,m
  CV ,m  R  CV ,m  3R 2  R 
 2.5R 1.5R  1.667
P1V11.667  P2V21.667
3
R
2
Recall:
C P ,m  CV ,m  R
Question 3
(b)
Calculate pressure at initial state first since it is not given;
P1  nRT1 V1
 1mol8.206 102 L  atm K  mol 300K  24.6L
49.21.667
 1.00atm
Calculate pressure at final state:
P1V11.667  P2V21.667
1.00atm24.61.667   P2 49.21.667


P2  1.00atm 24.61.667 49.21.667  0.315atm
Question 3
(b)
Calculate temperature at final state;
P2V  nRT2
T2  P2V nR

 0.315atm49.2L  1mol  8.206 102 L  atm / mol  K
 189K

Question 3
(c)
1.2
1
P (atm)
0.8
0.6
Adiabatic
0.4
Isothermal
0.2
0
0
20
40
V (L)
60
Question 4
One mole of liquid water at 30°C is adiabatically compressed,
P increasing from 1.00 to 10.00 atm. Since liquids and solids
are rather incompressible, it is fairly good approximation to
take V as unchanged for this process. Calculate q, ΔU, and ΔH
for this process.
Question 4
• Extract out all the information:
T1  30C
n  1mol
P1  1atm
P2  10atm
The process is adiabatic, ΔV=0
• Since the process is adiabatic,
q0
• Calculate work:
U  q  w
2
0
 0    PdV
0
1
Question 4
• Calculate ΔH
H  U  PV
H  U 2  U1   P2V2  P1V1 
 0  V P2  P1 
 0  V P2  P1 
V ?
mass
mass
n
1
MW
18 gmol 1
mass  18 g
Question 4
mass  18 g
mass
density 
volume
mass
volume 
density
density  1000 kg cm3
For water
18 g
1kg 106 cm3
3
volume 



18
cm
1000 kg cm3 1000 g 1m3
Question 4
H  V P2  P1 
 18cm3 10  1atm
 162cm3  atm  16.2 J