15.5.  Thermal Processes Using an Ideal Gas

ISOTHERMAL EXPANSION OR COMPRESSION

When a system performs work isothermally, the temperature remains constant. In Figure 15.9a, for instance, a metal cylinder contains n moles of an ideal gas, and the large mass of hot water maintains the cylinder and gas at a constant Kelvin temperature T. The piston is held in place initially so the volume of the gas is Vi. As the external force applied to the piston is reduced quasi-statically, the gas expands to the final volume Vf. Figure 15.9b gives a plot of pressure (P=nRT/V) versus volume for the process. The solid red line in the graph is called an isotherm (meaning “constant temperature”) and represents the relation between pressure and volume when the temperature is held constant. The work W done by the gas is not given by W=PDV=P(VfVi) because the pressure is not constant. Nevertheless, the work is equal to the area under the graph. The techniques of integral calculus lead to the following result for W:

 (15.3) 

Where does the energy for this work originate? Since the internal energy of any ideal gas is proportional to the Kelvin temperature ( for a monatomic ideal gas, for example), the internal energy remains constant throughout an isothermal process, and the change in internal energy is zero. The first law of thermodynamics becomes DU=0= QW. In other words, Q=W, and the energy for the work originates in the hot water. Heat flows into the gas from the water, as Figure 15.9a illustrates. If the gas is compressed isothermally, Equation 15.3 still applies, and heat flows out of the gas into the water. The following example deals with the isothermal expansion of an ideal gas.

(a) The ideal gas in the cylinder is expanding isothermally at temperature T. The force holding the piston in place is reduced slowly, so the expansion occurs quasi-statically. (b) The work done by the gas is given by the colored area.
Figure 15.9  (a) The ideal gas in the cylinder is expanding isothermally at temperature T. The force holding the piston in place is reduced slowly, so the expansion occurs quasi-statically. (b) The work done by the gas is given by the colored area.



Need more practice?
Need more practice?Interactive LearningWare 15.1

A bubble from the tank of a scuba diver contains 3.5×10–4 mol of gas. The bubble expands as it rises to the surface from a freshwater depth of 10.3 m. Assuming that the gas is an ideal gas and the temperature remains constant at 291 K, find the amount of heat that flows into the bubble.





Example 5  Isothermal Expansion of an Ideal Gas

Two moles of the monatomic gas argon expand isothermally at298 K, from an initial volume of Vi=0.025 m3 to a final volume of Vf=0.050 m3. Assuming that argon is an ideal gas, find (a) the work done by the gas, (b) the change in the internal energy of the gas, and (c) the heat supplied to the gas.

Reasoning and Solution

(a) The work done by the gas can be found from Equation 15.3:
(b) The internal energy of a monatomic ideal gas is (Equation 14.7) and does notchange when the temperature is constant. Therefore, .
(c) The heat Q supplied can be determined from the first law of thermodynamics:
 (15.1) 

ADIABATIC EXPANSION OR COMPRESSION

When a system performs work adiabatically, no heat flows into or out of the system. Figure 15.10a shows an arrangement in which n moles of an ideal gas do work under adiabatic conditions, expanding quasi-statically from an initial volume Vi to a final volume Vf. The arrangement is similar to that in Figure 15.9 for isothermal expansion. However, a different amount of work is done here, because the cylinder is now surrounded by insulating material that prevents the flow of heat, so Q=0 J. According to the first law of thermodynamics, the change in internal energy is DU=QW=W. Since the internal energy of an ideal monatomic gas is (Equation 14.7), it follows that where Ti and Tf are the initial and final Kelvin temperatures. With this substitution, the relation DU=W becomes

 (15.4) 

When an ideal gas expands adiabatically, it does positive work, so W is positive in Equation 15.4. Therefore, the term TiTf is also positive, so the final temperature of the gas must be less than the initial temperature. The internal energy of the gas is reduced to provide the necessary energy to do the work, and because the internal energy is proportional to the Kelvin temperature, the temperature decreases. Figure 15.10b shows a plot of pressure versus volume for an adiabatic process. The adiabatic curve (red) intersects the isotherms (blue) at the higher initial temperature [Ti=PiVi/(nR)] and the lower final temperature [Tf=PfVf/(nR)]. The colored area under the adiabatic curve represents the work done.

(a) The ideal gas in the cylinder is expanding adiabatically. The force holding the piston in place is reduced slowly, so the expansion occurs quasi-statically. (b) A plot of pressure versus volume yields the adiabatic curve shown in red, which intersects the isotherms (blue) at the initial temperature Ti and the final temperature Tf. The work done by the gas is given by the colored area.
Figure 15.10  (a) The ideal gas in the cylinder is expanding adiabatically. The force holding the piston in place is reduced slowly, so the expansion occurs quasi-statically. (b) A plot of pressure versus volume yields the adiabatic curve shown in red, which intersects the isotherms (blue) at the initial temperature Ti and the final temperature Tf. The work done by the gas is given by the colored area.

The reverse of an adiabatic expansion is an adiabatic compression (W is negative), and Equation 15.4 indicates that the final temperature exceeds the initial temperature. The energy provided by the agent doing the work increases the internal energy of the gas. As a result, the gas becomes hotter.

The equation that gives the adiabatic curve (red) between the initial pressure and volume (Pi, Vi) and the final pressure and volume (Pf, Vf) in Figure 15.10b can be derived using integral calculus. The result is

 (15.5) 

where the exponent g (Greek gamma) is the ratio of the specific heat capacities at constant pressure and constant volume, g=cP/cV. Equation 15.5 applies in conjunction with the ideal gas law, for each point on the adiabatic curve satisfies the relation PV=nRT.

Table 15.1 summarizes the work done in the four types of thermal processes that we have been considering. For each process it also shows how the first law of thermodynamics depends on the work and other variables.



Table 15.1   Summary of Thermal Processes
 Type of Thermal Process 
 Work Done 
 First Law of Thermodynamics (DU=QW
 Isobaric (constant pressure) 
 W=P(VfVi
  
 Isochoric (constant volume) 
 W=0
  
 Isothermal (constant temperature) 
  (for an ideal gas) 
  
 Adiabatic (no heat flow) 
  (for a monatomic ideal gas) 
  
  




Copyright © 2000-2003 by John Wiley & Sons, Inc. or related companies. All rights reserved.