Homework solutions
1. List the four ways in which the internal energy of a system can be changed.
heat (q) can flow in
heat can flow out
work (w)can be done on the system
work can be done by the system
- additional thought: what is the sign of )U in each case?
2. A system absorbs 6 kJ of heat, what is the change in the internal energy.
+ 6kJ
3. In compressing a system, 4.5 kJ of work is done on it, what is the change in the internal energy.
+4.5 kJ
4. A machine does 1000 kJ of work, what is the change in the internal energy of the machine?
Explain whether the sign should be positive or negative.
- 1000 kJ
3-1
Last lecture we started calorimetry and will continue to look at calorimetry at
constant volume (bomb calorimeter and constant pressure (like the lab expt.)
First we need to introduce ourselves to ENTHALPY (H): we will see that this
is heat flow at constant pressure and is easier to measure than U
)Usys = q sys+ wsys
work
Change in internal energy
Heat flow
+q : heat added to system from surroundings
-q : heat lost from system to surroundings
Watch the signs
+w : work done on system by surroundings
-w : work done by system on surroundings
3-2
PV work:
)V
P)V
expansion
+
+
-
compression -
-
+
w= -P)V
Enthalpy and Internal energy
we now need to be a bit more accurate about q and w
Constant Volume: At constant volume (bomb
calorimeter : sealed lid) heat added/lost is qv. At
constant volume work is ZERO (no volume change)
Constant Pressure: heat is called qp . Heat flow at
constant pressure is called ENTHALPY
-given the symbol H
3-3
So at constant pressure:
)U
= qp + w
)U
= )H - P)V
)H
= )U +P )V
Ebbing p 768
Good to remember these
• first let’s make sure we can
calculate w properly signs and
units
Examples:
Calculate w at 25OC and 1 atm pressure for
a) CO2 (g) + NaOH (s) ------> NaHCO3(s)
b) 2O3(g) -------> 3O2(g)
3-4
(a) we go from 1 mole of CO2 to zero moles of gas. Work is done on the
system by compressing it.
CO2 (g) + NaOH (s)
NaHCO3(s)
Work = P)V
)V = 24.5 L (Volume that was occupied by 1 mole of any gas at SATP)
or:
PV = nRT , so V = nRT/P = 1.00 x 8.314 x 298 / 1.01 x 102 = 24.5L
w = Pext)V= 24.5L atm.
- this must be expressed in SI units
1 atm in Pascals
(Nm-2)
This is done with the gas constant
R = 0.0821 L atm /K mole = 8.314 J/K mole
so 0.0821 L atm = 8.314 J
and 1 L atm = (8.314/0.0821)J = 101J
i.e., 24.5L atm = 24.5 x 101 J = +2.47 kJ
Useful to remember
that one
3-5
b) 2O3(g)
3O2(g)
here we go from 2 to 3 moles of gas. The system has to do work
on the surroundings
work is then -P)V. what is )V?
P)V = )nRT
)V = )nRT/P
= (1.00x 8.314 x 298)/101
= 24.5 L
work = -P)V = -1.00 x 24.5 L
= -24.5 Latm
= -2.47 kJ
3-6
Summary:
Energy : State function (internal Energy and Enthalpy) work and heat : path functions
expansion: work is -ve (-P)V = -)nRT): work is done by system therefore loses energy
•Specific Heat (symbol C)
• amount of heat required to raise 1g of material by 1 K (or 1 OC)
q = specific heat (JK-1g-1 ) x mass (g) x )T (K)
Heat Capacity (C)
•amount of heat required to raise temp of a body by 1 K (or 1 OC)
q = heat capacity (JK-1) x )T (K)
• Molar Heat Capacity (C)
•amount of heat required to raise temp of one mole of element or compound by 1 K (or 1
OC)
q = molar heat capacity (JK-1mol-1 ) x moles (n) x )T (K)
3-7
OK now we can do calorimetry properly: let’s measure )U
Isolated system
qsys = qcal + qchemicals = 0
qchemicals = - qcalorimeter
qcalorimeter = Ccal x )T
we have to relate q to )U ( and )H)
Example: 5.00g of a fuel were burned in a calorimeter holding 2000g of water. The heat
capacity of the reaction vessel was 1.84kJK-1 and the specific heat of water is 4.18 Jg-1K-1.
The temperature of the calorimeter rose by 5K. What is the sign and value of q?
Solution: heat energy is given to calorimeter and water:
q = 1.84 x 103 x 5
Energy to
bring vessel up
to temp
+ 2000 x 4.18 x 5
Energy to bring
water up to temp
Note:
q(cal) = - q(sys)
= 9200 + 41800 = 51.0 kJ
thus q sys = - 51.0 kJ
heat capacity of calorimeter (vessel + water) is
10.2 kJK-1 (i.e. q sys = 10.2 kJK-1 x -5) = - 51.0kJ
3-9
Heat flow measurements : two types of calorimetry and it is important to get
this straight right now
Constant volume
(bomb)
)U = q + w
Constant pressure
(coffee cup) Heat at const
)U = qp + w
press
w = 0 (P)V = 0)
qv = )U
---------------------
Experimentally
measured
to get )H
H = U + PV
)U = )H - P)V
To get )H
)H = qp
)H = )U +)P(V) +P)V
= )U +)nRT
-this is hard to see -- you
need to think about it
)P = )nRT/V
Heat at const press: easy to measure
3-10
Example : 1mol of CH4 (g) was burned with 2 mol of O2 (g) to form 1 mol CO2
(g) and 2 mol of H2O (l). 890.6 kJ of energy was released at 25OC and a constant
pressure of 1 atm. Calculate )H and )U.
CH4 (g) + 2 O2 (g)
is called a combustion reaction.
CO2 (g) +2 H2O(l)
qp = -890.6 kJ so )H = -890.6 kJ
we can get )U from PV = nRT:
)U = )H - P)V (see p 3-10)
)U = )H - )nRT
)U = )H - (nfinal - ninitial) x 8.314 x 298
)U = )H - (1-3) x 8.314 x 298
)U = -890.6 - (-4.96)
)U = -885.6 kJ
3-11