PHYSICAL CHEMISTRY IN BRIEF

CHAP. 3: FUNDAMENTALS OF THERMODYNAMICS [CONTENTS] 65

Solution

If the systemperformedwork, then, according to usage1.2,the workW =− 400 J. Hence the

change in the internal energy is

∆U=Q+W= 1000 + (−400) = 600 J.

If a system exchanges only reversible volume work with its surroundings during a thermo-

dynamic process [see4.1.1], then

dU= ̄dQ−pdV. (3.4)

If this process is isochoric, dV = 0. The change in the system’s internal energy then equals the

heat exchanged between the system and its surroundings

∆U=Q , [V]. (3.5)

It follows from the first law of thermodynamics that it is impossible to construct anyper-

petual motion machine of the first type—a hypothetical machine doing work cyclically

without receiving any energy from the surroundings. There is a simple proof of this fundamen-

tal assertion: since the first law of thermodynamics postulates that internal energy is a function

of state, we must have for a cyclic process

∮

dU= 0.

If a system does not exchange any heat with its surroundings ( ̄dQ= 0), we obtain from equation

(3.1)

W=

∮

dW= 0.

Hence, the work done is zero.
S Symbols: The symbol

∮

is used to denote a loop integral (the initial and the end point are

identical).

3.1.3 Second law of thermodynamics

There is a function of state calledentropyS.For its total differential dSwe write

dS =

̄dQ

T

, [reversible process], (3.6)