CH101 Spring 2012 Final Exam Review Questions Chapter 7
#1. A piece of iron of mass 72.4 g at 100. oC (373 K) is plunged into 100.0 g of
water initially at 10.0oC (283 K) in a Styrofoam cup. The final steady temperature
is 16.5oC (289.5 K). Determine the specific heat capacity of iron. The specific
heat capacity of water (Cp) is 4.18 J K-1 g-1.
Heat lost by iron (-ve) = Heat gained by water
Or - (mCpT)iron = (mCpT)water
i.e. (– 72.4 g) x Cp x (289.5 K– 373.0 K) = (100 g) x (4.18 J K-1 g-1) x (289.5 K–
283.0 K)
Or
Cp = (100 g) x (4.18 J K-1 g-1) x (6.5 K)
(72.4 g) x (83.5 K)
= 0.449 J K-1g-1
#2. A 6.00 L volume of air at a pressure of 1.00 atm is trapped in a cylinder
behind a sealed but moveable piston. A force is suddenly applied to the piston,
causing the volume in the cylinder to decrease to 5.45 L. Assuming ideal
behavior and constant temperature, calculated the work in joules.
The word “sudden” used here is meant to say that the constant external pressure
is equal to the final pressure of the gas during an irreversible expansion.
W = -P2(V2 – V1)
= -P2V2(1 –V1/V2)
= -P1V1(1 – V1/V2), since Boyle’s Law applies and we are not told P2 (Pext)
= -(1.00 atm)(6.00 L)(1 – 6.00 L/5.45 L)
= +0.606 L atm
= +61.3 J (1 L atm = 101.325 J)
#3. Adding a certain amount of heat to 100 g of water at constant pressure raises
its temperature by 3.79 o C. Adding the same amount of heat to 100 g of octane
at constant pressure raises its temperature by 9.63 oC. Find the specific heat (Cp)
of octane. [Cp of water = 4.18 J oC-1 g-1]
If qp(water) = qp(octane),
then m(water)Cp(water)T(water) = m(octane)Cp(octane) T(octane)
Since the masses are equal,
Cp(octane) = Cp(water) T(water)/ T(octane)
= (4.18 J g-1 oC-1) (3.79 oC)/9.63 oC)
= 1.65 J g-1 oC-1
#4. Calculate the heat and work associated with a process in which 5.00 mol of
gas expands reversibly at constant temperature T = 298 K from a pressure of 10
atm to 1.00 atm.
At constant T and n,
V2 = p1 = 10.0 (atm)
V1
p2
1.00 (atm)
= 10
Hence, since w = - n RT ln (V2/V1),
w = - 5.00 (mol) x 8.315 (J K-1 mol-1) x 298 (K) x ln 10
= - 28500 J
q = +28500 J (U = 0 for an isothermal process)
#5. Methanol (CH3OH(l)) obtained from biomass conversion is proposed to be
blended with gasoline (C8H18(l)) to extend our fossil-fuel reserves. Use the
following data to compare the enthalpy released per mole of fuel.
Hof values in kJ mol-1: (CH3OH(l)) = -238.4; C8H18(l) = -224.8; CO2(g) = -393.1;
H2O(l) = 285.5.
Firstly, the two equations required are:
CH3OH(l) + 3/2O2(g)

CO2(g) + 2H2O(l)
C8H18(l) + 25/2O2(g)

8CO2(g) + 9H2O(l)
CH3OH(l) + 3/2O2(g)
Hof (kJ mol-1)
-238.4

0
CO2(g) + 2H2O(l)
-393.1
2 x -285.5
Ho(reaction) = (-393.1) + [2 x (-285.5)] -(-238.4) - 0
= -725.7 kJ mol-1
C8H18(l) + 25/2O2(g)
Hof (kJ mol-1)
-224.8
0

8CO2(g) + 9H2O(l)
8 x (-393.1) 9 x (-285.5)
Ho(reaction) = [8 x(-393.1)] + [9 x (-285.5)] -(-224.8) = -5488 kJ mol-1
0
Gasoline is a more efficient fuel (per mole).
#6. An ideal monatomic gas (0.500 mol) at 1.00 atm and 273 K expands against
a constant pressure of 0.100 atm until the pressure reaches 0.200 atm and the
temperature reaches 210 K. Calculate the internal energy change, the work done
on the gas and the heat absorbed by the gas.
[ R = 0.0821 L atm K-1 mol-1, 8.315 J K-1 mol-1. 1 L atm = 101.325 J]
For an ideal monatomic gas Cv = 3/2R.
U = nCv T = 0.50(mol) x 3/2 x 8.315(J K-1 mol-1) x (-63 K)
= -393 J
w = -Pext V. Since V = V2 – V1, we can use PV = nRT to find both of these, the
initial and final states being known
For state 1, P1V1 = nRT1, V1 = 0.500(mol) x 0.0821(L atm K-1 mol-1) x 273(K) /
1.00(atm)
V1 = 11.2 L
For state 2, P2V2 = nRT2, V2 = 0.500(mol) x 0.0821(L atm K-1 mol-1) x 210(K) /
0.200 (atm)
V2 = 43.1 L
Hence V = 31.9 L
w = -0.100 x 31.9 = -3.19 Latm
= -323 J
Since U = q + w,
q = -393 + 323 J
= -70 J
#7. Use the Born-Haber cycle to calculate the lattice energy of sodium
chloride.
Hf o(NaCl) = -411.2 kJ; Hdiss(Cl2) = +121.7 kJ; Hvap (Na) = +107.3 kJ;
IE(Na) = +496 kJ; EA(Cl) = +349 kJ
Na+(g) + Cl(g)
-EA = -349 kJ
IE(Na) = +496 kJ
Na+(g) + Cl -(g)
H
Na(g) + Cl(g)
Hvap (Na) + Hdiss(Cl2)
= +107.3 kJ + 121.7 kJ
Na(s) + 1/2Cl2(g)
-HL
-Hf o(NaCl)
= +411.2 kJ
Na+Cl-(s)
HL = -Hf o(NaCl) + Hvap (Na) + Hdiss(Cl2) + IE(Na) -EA(Cl)
= 411.2 + 107.3 + 121.7 + 496 - 349 = +787 kJ