14.2. First Law of Thermodynamics http://www.ck12.org
What is the change in internal energy∆U?
Answer:
The internal energy of a gas can be expressed asU=^32 kNT, wherekis the Boltzmann constant,Nthe total number
of molecules (or atoms) andTthe temperature of the gas. For an isothermal process, the internal energy must remain
constant since the temperature remains constant, therefore∆U=0.
Illustrative Example 4
A gas-filled cylinder with a movable piston is put in contact with a heat reservoir so that the temperature of the gas
remains constant. If 400 J of heat are transferred out of the system, describe whether work was done by or on the
system.
Answer:
The sumQ+W=0 because∆U= 0 → 0 =Q+W→W=−Q. The work is equal to−Q, but since heat left the
systemQ<0, that is,Q=− 400 J→W=−(− 400 J) = + 400 J. Work is, therefore, done on the system.
Illustrative Example 5
a. An isochoric process transfers 1,500 J of heat to a gas inside a cylinder with a fixed piston. How much work is
done by the gas in the process?
Answer:
Since the process is isochoric, the volume of the gas must remain constant. If the gas does not expand or contract,
no work can be done by the gas on the environment or by the environment on the gas, since the displacement of the
piston is zero→W=P∆V=F∆x=F( 0 ) =0 (see example 14.1.1).
b. How much does the internal energy of gas change?
Answer:∆U=Q+W→∆U=Q+ 0 →∆U=Q→∆U= 1 , 500 J
Illustrative Example 6
Explain why a system may experience a temperature decrease during an adiabatic process.
Answer:
The heat cannot enter or leave the system(Q= 0 ). By the First Law, we have
∆U=Q+W= 0 +W→∆U=W.
If we assume that the system does work on the environment, thenW<0.
In cases when the work is less than zero, the internal energy of the system must decrease, since∆U=Uf−Ui<
0 →Uf<Ui.
The internal energy of a gas is directly related to the temperature of the gasU=^32 nRT. The smaller the internal
energyU, the cooler the temperatureT.