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Chapter 2 Reversible Isothermal Process in a Perfect Gas
The First Law of Thermodynamics Consider the special case of a reversible isothermal (constant-T) process in a perfect
gas. (Throughout this section, the system is assumed closed.) For a fixed amount of a
perfect gas, U depends only on T [Eq. (2.67)]. Therefore U 0 for an isothermal
change of state in a perfect gas. This also follows from dU C dT for a perfect gas.
V
The first law U q w becomes 0 q w and q w. Integration of dw
rev
PdV and use of PV nRT give
2 2 nRT 2 1
w P dV dV nRT dV nRT1ln V ln V 12
2
1 1 V 1 V
V 1 P 2
w q nRT ln nRT ln rev. isothermal proc., perf. gas (2.74)
V 2 P 1
where Boyle’s law was used. If the process is an expansion (V V ), then w (the work
1
2
done on the gas) is negative and q (the heat added to the gas) is positive. All the added
heat appears as work done by the gas, maintaining U as constant for the perfect gas.
It is best not to memorize an equation like (2.74), since it can be quickly derived from
dw PdV.
To carry out a reversible isothermal volume change in a gas, we imagine the gas to
System
be in a cylinder fitted with a frictionless piston. We place the cylinder in a very large
constant-temperature bath (Fig. 2.11) and change the external pressure on the piston at
an infinitesimal rate. If we increase the pressure, the gas is slowly compressed. The work
Bath done on it will transfer energy to the gas and will tend to increase its temperature at an
infinitesimal rate. This infinitesimal temperature increase will cause heat to flow out of
the gas to the surrounding bath, thereby maintaining the gas at an essentially constant
Figure 2.11
temperature. If we decrease the pressure, the gas slowly expands, thereby doing work
Setup for an isothermal volume on its surroundings, and the resulting infinitesimal drop in gas temperature will cause
change. heat to flow into the gas from the bath, maintaining constant temperature in the gas.

EXAMPLE 2.5 Calculation of q, w, and U

A cylinder fitted with a frictionless piston contains 3.00 mol of He gas at P
1.00 atm and is in a large constant-temperature bath at 400 K. The pressure is re-
versibly increased to 5.00 atm. Find w, q, and U for this process.
It is an excellent approximation to consider the helium as a perfect gas.
Since T is constant, U is zero [Eq. (2.68)]. Equation (2.74) gives
1
1
w 13.00 mol218.314 J mol K 21400 K2 ln 15.00>1.002 19980 J2 ln 5.00
4
w 19980 J211.6092 1.61 10 J
4
Also, q w 1.61 10 J. Of course, w (the work done on the gas) is pos-
itive for the compression. The heat q is negative because heat must flow from the
gas to the surrounding constant-temperature bath to maintain the gas at 400 K as
it is compressed.
Exercise
0.100 mol of a perfect gas with C 1.50R expands reversibly and isother-
V,m
mally at 300 K from 1.00 to 3.00 L. Find q, w, and U for this process. (Answer:
274 J, 274 J, 0.)

Reversible Constant-P (or Constant-V) Process in a Perfect Gas
The calculations of q, w, and U for these processes were shown in Example 2.4.