4.6 Entropy (Page 3/5)

University physics volume 2 Page 3 / 5

Stirling engine

The steps of a reversible Stirling engine are as follows. For this problem, we will use 0.0010 mol of a monatomic gas that starts at a temperature of

133 ° C

and a volume of

0.10 m^{3}

, which will be called point A . Then it goes through the following steps:

Step AB : isothermal expansion at $133 ° C$ from $0.10 m^{3}$ to $0.20 m^{3}$
Step BC : isochoric cooling to $33 ° C$
Step CD : isothermal compression at $33 ° C$ from $0.20 m^{3}$ to $0.10 m^{3}$
Step DA : isochoric heating back to $133 ° C$ and $0.10 m^{3}$

(a) Draw the pV diagram for the Stirling engine with proper labels.

(b) Fill in the following table.

Step	W (J)	Q (J)	$Δ S$ (J/K)
Step AB
Step BC
Step CD
Step DA
Complete cycle

(c) How does the efficiency of the Stirling engine compare to the Carnot engine working within the same two heat reservoirs?

Strategy

Using the ideal gas law, calculate the pressure at each point so that they can be labeled on the pV diagram. Isothermal work is calculated using

W = n R T ln (\frac{V_{2}}{V_{1}}),

and an isochoric process has no work done. The heat flow is calculated from the first law of thermodynamics,

Q = Δ E_{int} - W

where

Δ E_{int} = \frac{3}{2} n R Δ T

for monatomic gasses. Isothermal steps have a change in entropy of Q / T , whereas isochoric steps have

Δ S = \frac{3}{2} n R ln (\frac{T_{2}}{T_{1}}) .

The efficiency of a heat engine is calculated by using

e_{Stir} = W / Q_{h} .

Solution

The graph is shown below.

The completed table is shown below.

Step	W (J)	Q (J)	$Δ S$ (J/K)
Step AB Isotherm	2.3	2.3	0.0057
Step BC Isochoric	0	–1.2	0.0035
Step CD Isotherm	–1.8	–1.8	–0.0059
Step DA Isochoric	0	1.2	–0.0035
Complete cycle	0.5	0.5	~ 0

The efficiency of the Stirling heat engine is

$e_{Stir} = W / Q_{h} = (Q_{A B} + Q_{C D}) / (Q_{A B} + Q_{D A}) = 0.5 / 4.5 = 0.11 .$

If this were a Carnot engine operating between the same heat reservoirs, its efficiency would be

e_{Car} = 1 - (\frac{T_{c}}{T_{h}}) = 0.25 .

Therefore, the Carnot engine would have a greater efficiency than the Stirling engine.

Significance

In the early days of steam engines, accidents would occur due to the high pressure of the steam in the boiler. Robert Stirling developed an engine in 1816 that did not use steam and therefore was safer. The Stirling engine was commonly used in the nineteenth century, but developments in steam and internal combustion engines have made it difficult to broaden the use of the Stirling engine.

The Stirling engine uses compressed air as the working substance, which passes back and forth between two chambers with a porous plug, called the regenerator, which is made of material that does not conduct heat as well. In two of the steps, pistons in the two chambers move in phase.

Summary

The change in entropy for a reversible process at constant temperature is equal to the heat divided by the temperature. The entropy change of a system under a reversible process is given by $Δ S = \int_{A}^{B} d Q / T$ .
A system’s change in entropy between two states is independent of the reversible thermodynamic path taken by the system when it makes a transition between the states.

Conceptual questions

Does the entropy increase for a Carnot engine for each cycle?

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Is it possible for a system to have an entropy change if it neither absorbs nor emits heat during a reversible transition? What happens if the process is irreversible?

Entropy will not change if it is a reversible transition but will change if the process is irreversible.

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Problems

Two hundred joules of heat are removed from a heat reservoir at a temperature of 200 K. What is the entropy change of the reservoir?

–1 J/K

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In an isothermal reversible expansion at $27 °C$ , an ideal gas does 20 J of work. What is the entropy change of the gas?

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An ideal gas at 300 K is compressed isothermally to one-fifth its original volume. Determine the entropy change per mole of the gas.

–13 J(K mole)

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What is the entropy change of 10 g of steam at $100 °C$ when it condenses to water at the same temperature?

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A metal rod is used to conduct heat between two reservoirs at temperatures $T_{h} and T_{c},$ respectively. When an amount of heat Q flows through the rod from the hot to the cold reservoir, what is the net entropy change of the rod, the hot reservoir, the cold reservoir, and the universe?

$- \frac{Q}{T_{h}}, \frac{Q}{T_{c}}, Q (\frac{1}{T_{c}} - \frac{1}{T_{h}})$

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For the Carnot cycle of [link] , what is the entropy change of the hot reservoir, the cold reservoir, and the universe?

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A 5.0-kg piece of lead at a temperature of $600 °C$ is placed in a lake whose temperature is $15 °C$ . Determine the entropy change of (a) the lead piece, (b) the lake, and (c) the universe.

a. –540 J/K; b. 1600 J/K; c. 1100 J/K

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One mole of an ideal gas doubles its volume in a reversible isothermal expansion. (a) What is the change in entropy of the gas? (b) If 1500 J of heat are added in this process, what is the temperature of the gas?

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One mole of an ideal monatomic gas is confined to a rigid container. When heat is added reversibly to the gas, its temperature changes from $T_{1} to T_{2} .$ (a) How much heat is added? (b) What is the change in entropy of the gas?

a. $Q = n R Δ T$ ; b. $S = n R ln (T_{2} / T_{1})$

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(a) A 5.0-kg rock at a temperature of $20 °C$ is dropped into a shallow lake also at $20 °C$ from a height of $1.0 \times 10^{3} m$ . What is the resulting change in entropy of the universe? (b) If the temperature of the rock is $100 °C$ when it is dropped, what is the change of entropy of the universe? Assume that air friction is negligible (not a good assumption) and that $c = 860 J/kg \cdot K$ is the specific heat of the rock.

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Source: OpenStax, University physics volume 2. OpenStax CNX. Oct 06, 2016 Download for free at http://cnx.org/content/col12074/1.3

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