Page 44 - [B._MURPHY,_C._MURPHY,_B._HATHAWAY]_A_working_meth

28 Chapter 3
Solution:
= mcAT = mc(Tfina1 - Tinitial) = (500 x 4.184) x (348-298) =
104600 J = 104.6 kJ. Notice that the sign of q is +ve. This means
that the water has absorbed heat.
1 Example No. 2: Show that for an ideal gas, Cm,p = Cm,v + R 1

Proo$ At constant pressure: AH = AU + pAV
+ nCm,,AT = nCm,vAT + pAV = nCm,,AT + nRAT, since from
the equation of state of an ideal gas, PA V = nRAT.
But, for 1 mole of an ideal gas, n = 1.
+ (l)Cm,pAT = (l)Cm,vAT + (1)RAT. Then, dividing across by AT,
Cm,p = Cm,v + R.
In the above expression, Cm,p has to be greater than Cm,v, since a
certain amount of heat is needed to increase the temperature at
constant volume. However, at constant pressure, more heat is required
to raise the temperature and do the work, expanding the volume of
the gas (Chapter l-Charles' Law: V oc T, for a fixed mass of gas at
constant pressure).

Example on Heat Capacities

Example: C (Au) = 25.4 J K-' mol-' and Cm,p(HzO) = 75.3
"iP
J K-' mol- . If a gold stud initially at a room temperature of
25 "C is dropped into 12 g of water at 8 "C, the final temperature of
the water reaches 13 "C. Given that the molar masses of Au, H and
0 are 197, 1 and 16 g mol- respectively, determine the mass of the
stud.

Solution:
1. Read the question carefully.
2. Identify the data:
M(H20) = 2(1) + 16 = 18 g mol-'; M(Au) = 197 g mol-';
12 g of H20
C,,p(Au) = 25.4 J K-' mol-'; Cm,,(H20) = 75.3 J K-' mol-';
Tinitial(Au) (K) = T("C) + 273 = (25 + 273) K = 298 K;
Tfin,l(Au) (K) = (13 + 273) K = 286 K;
Tinitial(H20) (K) = (8 + 273) = 281 K;
Tfinal(HZO) (K) = (13 + 273) = 286 K.