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10 Heat and Matter
10.52 (a) The internal energy of an ideal gas is given by U = nC V T . In isothermal
expansion where the temperature and the amount of gas remain constant,
the internal energy does not change. Thus U = 0.
(b) The work done is
W = −
V 2
V 1
PdV = −
V 2
V 1
nRT
V
dV
= −nRT
V 2
V 1
dV
V
= −nRT ln(V 2 /V 1 ).
(c) Using the first law of thermodynamics, Q = U + W , and putting
U = 0, we have
Q = W = nRT ln(V 2 /V 1 )
(d) The change in entropy is
S =
ds =
dQ
T
=
1
T
dQ
because the temperature T does not change. Thus
S = n R ln(V 2 /V 1 )
(e) By assumption the temperature of the reservoir does not change and
because it loses heat Q to the gas, the entropy change of the reservoir
will be
S res = −S = −n R ln(V 2 /V 1 )
Therefore the entropy change of the system plus the reservoir equals
zero, which is the definition of a reversible process.
10.53 (a) The internal energy U of a system tends to increase if energy is added
as heat Q and tends to decrease if energy is lost as work W done by the
system.
Heat is energy that is transferred from one body to another due to difference in temperature of the bodies.
Enthalpy (H ) is the total heat and is defined by
H = U + PV