{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

lecture 7

# lecture 7 - The 2nd Law For any equilibrium(=...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: The 2nd Law For any equilibrium (= reversible) process the quantity dS = δq/T is an exact differential (i.e. an increment of a function of state). In other words, = S(2) – S(1) where S is a function of state. It is called entropy. Remember that q is not a function of state. The 2nd law further postulates that for an irreversible process the entropy satisfies the inequality dS > δq/T Therefore for an isolated system that does not exchange heat with its environment the entropy can only increase or remain constant. Examples of using the 2nd law Entropy on an ideal gas. First, let us convince ourselves that entropy is a function of state for an ideal gas. This will also provide us with a useful formula for the entropy of an ideal gas that we will be using extensively in the future. We have δq = dU + P dV or δq/T = dU/T + P dV/T Now recall that dU = n δq/T = dT and that T = (PV)/(Rn), which gives: n dT/T + R n dV/V Assuming, for simplicity, that S = δq/T = is independent of T, we get: n ln T + R n ln V + constant Similarly, we can derive an expression for the entropy of an ideal gas as a function of P and T. Start with and write P dV = R n dT – V dP, which gives us: δq/T = ( or S = n ln T  ­R n ln P + constant and . To write a logarithm of a quantity that is not +R) n dT/T  ­ R n dP/P if we recall the connection between dimensionless (P or V) is a bad habit; however since we are interested not in the absolute value of S but in its change from state 1 to state 2, we will always be dealing with quantities like ln P(2) – ln P(1) = ln (P(2)/P(1)), which are ok to use. Calculating the entropy change in a constant pressure or constant volume process: If heating is done reversibly (i.e., the heater temperature is always just slightly above the temperature of the system) In a process where T goes from T1 to T2, Finally, if the heat capacity does not depend on T then this becomes Consequences of the 2nd law (aka other formulations of the 2nd law): Consequence 1. Heat does not spontaneously flow from a colder object to a hotter one. Proof (sort of). Suppose, e.g., that we have spontaneous flow δq > 0 of heat from a cold object to a hot object at constant pressure. Then we will have . But this would contradict the 2nd law since for a system (hot object + cold object) that does not exchange heat with the rest of the world the entropy must increase in a spontaneous process. Consequence 2. It is impossible to take heat q from a reservoir, convert it all into work, and return into the original state (no cyclic heat engine with the efficiency equal to 1). Otherwise we could extract heat from a colder object, convert it all to work (e.g. left a weight), then let the weight fall on a hotter object heating it up. The net effect of all this would be transfer of heat from a hotter object to a colder object, which would violate Consequence 1. Notice that the requirement of being cyclic is essential. You can extract heat from a thermal reservoir to produce work by having isothermal expansion of a gas that is in contact with the reservoir. However if you want to reuse the gas it will have to return to its original state. If you compress it again at the same temperature, you will have to do work on the gas. The net effect of expansion and compression (assuming both processes performed in a reversible fashion) would be that that gas has done zero work. Consequence 3. Entropy of the world increases (if we assume he world to be an isolated system) ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online