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Chapter 5 8 Chapter 5 Ideal Gas 5 Equation of state of a gas A gas is the thermometric substance because the ratio of the pressure P of a gas at any temperature to the pressure PTP of the same gas at the triple point, as both P and PTP approach zero, approaches a value independent of the nature of t...

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Chapter 5 – 8 Chapter 5 Ideal Gas 5.1 Equation of state of a gas •

•

•

A gas is the best-behaved thermometric substance because the ratio of the pressure P of a gas at any temperature to the pressure PTP of the same gas at the triple point, as both P and PTP approach zero, approaches a value independent of the nature of the gas o The limiting value of this ratio, multiplied by 273.16K, was defined to be the ideal-gas temperature T of the system at whose temperature the gas exerts the pressure P The relation between Pv and P may be expressed for a real gas by means of a power series (or virial expansion) of the form 𝑃𝑣 = 𝐴(1 + 𝐵𝑃 + 𝐶𝑃2 + ⋯ ) …(5.1) o A, B, C, etc. are the virial coefficients and depends on the temperature and on the nature of the gas ▪ In the pressure range from 0 to about 40 standard atmospheres, the relation between Pv and P is practically linear, so that only the first two terms in the expansion are significant ▪ The greater the pressure range, the larger the number of terms in the virial expansion The first virial coefficient A is independent of the nature of the gas and depends only on 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑜𝑛𝑙𝑦 temperature. Thus, lim (𝑃𝑣) = 𝐴 = { 𝑖𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡 𝑜𝑓 𝑔𝑎𝑠 𝑃→0

5.2 Internal energy of a real gas •

• •

Imagine a thermally insulated vessel with rigid walls, divided into two compartments by a partition. Suppose that there is a gas in one compartment and that the other contains a vacuum. If the partition is removed, the gas will undergo adiabatic free expansion (Joule expansion) in which no work is done and no heat is transferred o From first law, since Q and W are zero, it follows that the internal energy remains unchanged during a free expansion The internal energy of any gas is a function of any two of the coordinates P, V, and T If no temperature change takes place in a free expansion of a gas, U is independent of V and of P, and therefore, U is a function of T only

5.3 Ideal gas • • • •

Ideal gas has properties that do not correspond to those of any existing gas, but are approximately those of a real gas at low pressures A real gas at pressures below about twice standard atmospheric pressure may be treated as the ideal gas without introducing an error greater than a few percent Even in the case of a saturated vapor in equilibrium with its liquid, the ideal-gas equation of state may be used with only a small error if the vapor pressure is low For an infinitesimal quasi-static process of a hydrostatic system, the first law is 𝑑𝑄 = 𝑑𝑈 + 𝑃𝑑𝑉 𝜕𝑈

and the heat capacity at constant volume is 𝐶𝑉 = ( 𝜕𝑇 )

𝑉

•

For ideal gases, U is a function of T only, so the partial derivative with respect to T is the same as 𝑑𝑈

the total derivative: 𝐶𝑉 = 𝑑𝑇 o 𝑑𝑄 = 𝐶𝑉 𝑑𝑇 + 𝑃𝑑𝑉 …(5.8) o For an infinitesimal quasi-static process, 𝑃𝑑𝑉 + 𝑉𝑑𝑃 = 𝑛𝑅𝑑𝑇 o Substitute the above equation into 5.8: 𝑑𝑄 = (𝐶𝑉 + 𝑛𝑅)𝑑𝑇 − 𝑉𝑑𝑃 o

• •

•

Divide by dT:

𝑑𝑄

𝑑𝑇

= 𝐶𝑉 + 𝑛𝑅 −

𝑑𝑃 𝑉 𝑑𝑇

…***

At constant pressure, the left-hand member becomes CP and dP = 0 → 𝐶𝑃 = 𝐶𝑉 + 𝑛𝑅 …(5.9) CP is larger than CV because as heat is supplied to a system at constant pressure, the gas expands and works against the external pressure, which is equal to the pressure of the gas in a quasistatic process. Thus, CP includes work of expansion, which is not found in CV Since ***, we find 𝑑𝑄 = 𝐶𝑃 𝑑𝑇 − 𝑉𝑑𝑃 …(5.10)

5.4 Experimental determination of heat capacities • •

•

The results of the measurements of CV and CP on gases at low pressures (ideal gases) can be stated in a simple manner in terms of molar heat capacities All ideal gases: o cV is a function of T only o cP is a function of T only, and is greater than cV o cP – cV is not a function of T, but equal to R 𝑐 o the ratio 𝑃 = 𝛾 is a function of T only, and is greater than 1 𝑐 𝑉

Monatomic gases, such as He, Ne, and A, and most metallic vapors, such as the vapors of Na, Cd, and Hg: o o o

•

3

cV is constant over a wide temperature range and is very nearly equal to 𝑅 2 5

cP is constant over a wide temperature range and is very nearly equal to 𝑅 2 the ratio

𝑐𝑃

𝑐𝑉

= 𝛾 is constant over a wide temperature range and is very nearly equal to

So-called permanent diatomic gases, namely, air, H2, D2, O2, N2, NO, and CO: o

5

cV is constant at ordinary temperatures, being equal to about 𝑅, and increases as the 2 temperature is raised

•

5 3

7 𝑅, 2

o

cP is constant at ordinary temperatures, being equal to about

o

temperature is raised 𝑐 7 the ratio 𝑃 = 𝛾 is constant at ordinary temperatures, being equal to about , and

and increases as the 3

𝑐𝑉

decreases at the temperature is raised Polyatomic gases and gases that are chemically active, such as CO2, NH3, CH4, Cl2, and Br2 o cP, cV, and 𝑐𝑃 /𝑐𝑉 vary with the temperature, the variation being different for each gas

5.5 Quasi-static adiabatic process • •

Start with equations 5.8 and 5.10 In adiabatic process, dQ = 0 → 𝑉𝑑𝑃 = 𝐶𝑃 𝑑𝑇 and 𝑃𝑑𝑉 = −𝐶𝑉 𝑑𝑇

•

Divide first equation by second:

𝑑𝑃 𝑃

𝐶 𝑑𝑉 𝑉 𝑉

= − 𝐶𝑃

→

𝑑𝑃 𝑃

= −𝛾

𝑑𝑉 𝑉

• • • •

The equation of state 𝑃𝑉 𝛾 = 𝑐𝑜𝑛𝑠𝑡. holds at all equilibrium states through which the ideal gas passes during a quasi-static adiabatic process …(5.11) A free expansion is an adiabatic process but is not quasi-static, because the gas rushing into the vacuum passes through nonequilibrium states before finally achieving equilibrium Equation 5.11 cannot be applied to states traversed by the ideal gas during a free expansion or any adiabatic process that is not quasi-static 𝜕𝑃

Slope of any adiabatic curve is ( 𝜕𝑉) = −𝛾 o

𝑆

𝑃 𝑉

…(5.12)

S denotes a reversible adiabatic process

5.6 Rüchhardt’s method of measuring γ

•

The gas is contained in a large jar of volume V. Fitted to the jar is a glass tube with an accurate bore of cross-sectional area A, into which a metal ball of mass m fits snugly like a piston. Since the gas is slightly compressed by the steel ball in its equilibrium position, its pressure P is slightly 𝑚𝑔 larger than atmospheric pressure P0. Thus, neglecting friction, 𝑃 = 𝑃0 +

•

If the ball is given a slight downward displacement and then let go, it will oscillate with a period τ. Friction will cause the ball to come to rest eventually. Let the displacement of the ball from its equilibrium position at any moment be denoted by y, where y is positive when the ball is above the equilibrium position and negative below. A small positive displacement causes an increase in volume which is very small compared with the equilibrium volume V and which, therefore, can be denoted by dV, where 𝑑𝑉 = 𝑦𝐴 Similarly, a small positive displacement causes a decrease in pressure which is very small compared with the equilibrium pressure P and which, therefore, can be denoted by dP, where

•

𝐴

dP is a negative quantity. The resultant force F acting on the ball is equal to AdP if we neglect 𝐹

• •

friction, or 𝑑𝑃 = 𝐴

Notice that, when y is positive, dP is negative and, therefore, F is negative → F is a restoring force As the ball oscillates fairly rapidly, the variations of P and V are adiabatic, because there is not enough time for appreciable heat transfer. Since the variations are also quite small, the states through which the gas passes can be considered to be approximately states of equilibrium. Therefore, we may assume that the changes of P and V represent an approximately quasi-static adiabatic process, and we may write 𝑃𝑉 𝛾 = 𝑐𝑜𝑛𝑠𝑡. and 𝛾𝑃𝑉 𝛾−1 𝑑𝑉 + 𝑉 𝛾 𝑑𝑃 = 0 o o

Substituting for dV and dP, we get 𝐹 = −

𝛾𝑃𝐴2 𝑉

𝑦

This equation expresses the fact that the restoring force is directly proportional to the displacement and is in the opposite direction, which is Hooke’s law. This is precisely the 𝑚

condition for simple harmonic motion, for which the period τ is 𝜏 = 2𝜋√ −𝐹/𝑦 o •

Consequently, 𝜏 = 2𝜋√

𝑚𝑉

𝛾𝑃𝐴2

, and as a result, 𝛾 =

4𝜋2 𝑚𝑉 𝐴2 𝑃𝜏2

…(5.14)

Rüchhardt’s method involves errors due to three simplifying assumptions: o The gas is ideal o There is no friction (this assumption is responsible for the largest error, ~3%) o Volume changes are strictly adiabatic

5.7 Velocity of a longitudinal wave

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Consider a gas enclosed in a cylinder and held in place by a piston exerting pressure P. If a compression is produced by moving the piston to the right at constant velocity, w0, the wave front of the pressure pulse will travel with a different constant velocity w, depending on the properties of the gas The “free body” for Newton’s second law is volume V of the gas whose initial uncompressed length is wt and whose uncompressed volume 𝑉 = 𝐴𝑤𝑡, where A is the cross-sectional area of the cylinder. If ρ if the density of the normal or uncompressed gas, the mass of the free body is ρAwt

•

Let us suppose that the piston shown in the lower part exerts a force 𝐴(𝑃 + Δ𝑃) as the piston moves to the right with a constant velocity w0. The compression moves with constant velocity w, so that in time t the compression has traveled a distance wt while the piston has traveled a distance w0t. At time t, rate of increase of mass of the compressed column =

• • • • • • •

𝜌𝐴𝑤𝑡 𝑡

The entire compressed column has a velocity w0 equal to that of the piston. Therefore, rate of increase of momentum of the compressed column = 𝜌𝐴𝑤𝑤0 The free body is acted on by a force 𝐴(𝑃 + Δ𝑃) to the right and a force AP to the left. Therefore, unbalanced force on the compressed column = 𝐴 Δ𝑃 From Newton’s second law, the unbalanced force is equal to the rate of change of momentum, 𝑤 𝐴Δ𝑃 = 𝜌𝐴𝑤𝑤0 → Δ𝑃 = 𝜌𝑤 2 0 𝑤

The “uncompressed free body” of volume V = Awt has undergone a compression (−Δ𝑉) = 𝐴𝑤𝑜 𝑡. That is, −

Δ𝑉

Therefore, Δ𝑃 =

𝑤 𝐴𝑤0 𝑡 = 𝑤0 𝐴𝑤𝑡 Δ𝑉 𝜌𝑤 2 (− 𝑉 ) → 𝑤 2 𝑉

=

=

−1

…(5.15)

𝜌 Δ𝑉 ( ) 𝑉 Δ𝑃

In view of the properties of ordinary matter, the volume changes which take place under the influence of a longitudinal wave at ordinary frequencies are adiabatic, not isothermal 1

Δ𝑉

Returning to 5.15 and identifying −( 𝑉)( ) as the reversible adiabatic compressibility 𝜅𝑆 , we Δ𝑃 1

have 𝑤 2 = 𝜌𝜅 …(5.16) 𝑆

• •

= 𝜌𝐴𝑤

1

𝜕𝑉

The adiabatic compressibility can be calculated for the ideal gas used 5.12: 𝜅𝑆 = − 𝑉 ( 𝜕𝑃) =

Density is 𝜌 =

𝑀 𝑣

→ 5.16 becomes 𝑤 2 =

𝛾𝑃𝑣 𝑀

=

𝛾𝑅𝑇 𝑀

…(5.17)

…𝛾 =

𝑀𝑤 2 𝑅𝑇

𝑆

1

𝛾𝑃

Chapter 6 6.1 Conversion of work into heat and vice versa •

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•

•

Work of any kind W may be done on a system in contact with a reservoir, causing heat Q to leave the system without altering the state of the system. The system acts merely as an intermediary To study the conversion of heat into work, a series of processes is needed in which a system is brought back to its initial state (a cycle) Each of the processes that constitute a cycle involves either the performance of work or a flow of heat between the system and its surroundings, which consist of a heat reservoir at a higher temperature than the system (a “high-temperature reservoir”) and a heat reservoir at a lower temperature than the system (a “low-temperature reservoir”) For one complete cycle, let o |QH| represent the heat exchanged between the high-temperature reservoir and the system o |QL| represent the heat exchanged between the low-temperature reservoir and the system o |W| represent the work exchanged between the system and the surroundings If |QH| is larger than |QL| and if |W| is done by the system, then the machine that causes the system to undergo the cycle is a heat engine

The purpose of a heat engine is to deliver work continuously to the surroundings by performing the same cycle over and over again The net work in the cycle is the output, and the heat absorbed from the high-temperature reservoir by the system is the input o

• • •

The thermal efficiency of the engine is defined as 𝜂 =

•

•

…(6.1)

Apply first law and remember there is no change of internal energy: |𝑄𝐻 | − |𝑄𝐿 | = |𝑊| o

•

|𝑊| |𝑄𝐻 |

Therefore, 𝜂 =

|𝑄𝐻 |−|𝑄𝐿 | |𝑄𝐻 |

= 1−

|𝑄𝐿 |

|𝑄𝐻 |

…(6.2)

If an engine could be built to operate in a cycle in which there is no outflow of heat from the working substance to the low-temperature reservoir, then there would be 100% conversion of heat from the high-temperature reservoir into work. But there must always be an outflow of heat from an engine, so the efficiency of a heat engine is always less than 100% The transformation of heat into work is usually accomplished by the internal-combustion engine (gasoline engine and diesel engine) and the external-combustion engine (steam engine and Stirling engine) In both types of heat engine, a gas or mixture of gases is contained in the space between a cylinder, closed at one end, and a piston. The gas in the confined space is the system, which undergoes a cycle, thereby causing a reciprocating piston to impart a motion of rotation to a shaft, which acts against an opposing force. It is necessary, in all engines, that the gas in the confined space, at some time in the cycle, be raised to a high temperature and a high pressure, the pressure providing the force that performs external work

6.2 Gasoline engine •

• • •

The cycle involves the performance of six processes, four of which require vertical motion of the piston and are called strokes: o Intake stroke o Compression stroke o Combustion o Power stroke o Exhaust o Exhaust stroke Friction, turbulence, loss of heat by conduction, and the chemical reaction between gasoline vapor and oxygen render an exact mathematical analysis quite difficult A drastic but useful simplification that neglects these troublesome effects creates an idealized gasoline engine that performs the Otto cycle The behavior of a gasoline engine can be approximated by assuming a set of ideal conditions as follows: o The working substance is at all times air, which behaves like an ideal gas with constant heat capacities o All processes are quasi-static o There is no friction or turbulence o There is no loss of heat through the walls of the combustion chamber

o

•

•

•

•

•

• •

The processes are reversible

Process 5 → 1: quasi-static intake stroke, isobaric at atmospheric pressure, volume of the combustion chamber varies from zero to V1 as the number of moles varies from zero to n, according to 𝑃0 𝑉 = 𝑛𝑅𝑇1 , where P0 is the atmospheric pressure and T1 is the temperature of the outside air Process 1 → 2: quasi-static adiabatic compression stroke, no friction, no loss of heat through the cylinder wall, temperature rises from T1 to T2, according to 𝑇1 𝑉1𝛾−1 = 𝑇2 𝑉2𝛾−1 where V1 is the larger volume when the piston is at the bottom of the compression stroke and V2 the smaller volume when the piston is at the top Process 2 → 3: quasi-static isochoric increase of temperature and pressure of n moles of air, imagined to be brought about by an absorption of heat |QH| from a series of external hightemperature reservoirs whose temperature range from T2 to T3. This process is meant to approximate the effect of the combustion in a gasoline engine when the piston is essentially motionless at the top of the stroke Process 3 → 4: quasi-static adiabatic power stroke, drop in temperature from T3 to T4, according to 𝑇3 𝑉2𝛾−1 = 𝑇4 𝑉1𝛾−1 , where V1 is larger than V2. This process represents the power stroke Process 4 → 1: quasi-static isochoric drop in temperature and pressure of n moles of air, brought about by a rejection of heat |QL| to a series of low-temperature external reservoirs ranging in temperature from T4 to T1. This process is meant to approximate the drop to atmospheric pressure upon opening the exhaust valve, but in reality, the temperature does not actually drop to the temperature of the outside air as it leaves the exhaust port Process 1 → 5: quasi-static exhaust stroke, isobaric at atmospheric pressure… Actual operating thermal efficiency of a gasoline engine is 20 – 30%

6.3 Diesel engine

•

Idealized diesel engine that performs Diesel cycle

6.6 Heat engine; Kelvin-Planck statement of the second law •

Important characteristics of heat-engine cycles: o There is some process or series of processes during which there is an absorption of heat from an external reservoir at a higher temperature o There is some process or series of processes during which heat is rejected to an external reservoir at a lower temperature

•

No heat engine has been developed that converts the heat extracted from a reservoir at a higher temperature into work without rejecting some heat to a reservoir at a lower temperature. This statement constitutes the second law of thermodynamics Kelvin-Planck statement of the second law: It is impossible to construct an engine that, operating in a cycle, will produce no effect other than the extraction of heat from a reservoir and the performance of an equivalent amount of work The continuous operation of a machine that creates its own energy and thus violates the first law is a perpetual motion machine of the first kind The continuous operation of a machine that utilizes the internal energy of only one heat reservoir, thus violating the second law, is a perpetual motion machine of the second kind

•

• •

Chapter 8 8.1 Reversible part of the second law

•

• • •

Consider a reversible process represented by i → f on the diagram above. The nature of the system is not essential. The dashed curves through i and f, respectively, represent portions of adiabatic processes. Let us draw a curve, a → b, representing an isothermal process, in such a way that the area under the smooth curve if is qual to the area under the zigzag sequence of processes, path iabf. Then, the work done in traversing both paths is the same, or 𝑊𝑖𝑓 = 𝑊𝑖𝑎𝑏𝑓 From the first law, 𝑄𝑖𝑓 = 𝑈𝑓 − 𝑈𝑖 − 𝑊𝑖𝑓 and 𝑄𝑖𝑎𝑏𝑓 = 𝑈𝑓 − 𝑈𝑖 − 𝑊𝑖𝑎𝑏𝑓 o Therefore 𝑄𝑖𝑓 = 𝑄𝑖𝑎𝑏𝑓 Since no heat is transferred in the two adiabatic processes ia and bf → 𝑄𝑖𝑓 = 𝑄𝑎𝑏 …(8.1) If we are given a reversible process in which the temperature may change in any manner, it is always possible to find a reversible zigzag path between the same two states, consisting of an adiabatic process followed by an isothermal process followed by an adiabatic process, such that the heat transferred during the isothermal segment is the same as that transferred during the original process

•

Consider the smooth closed curve on the generalized work diagram above. Since no two adiabatic lines can intersect, a number of adiabatic lines may be drawn, dividing the cycle into a number of adjacent strips. A zigzag closed path may now be drawn, consisting of alternate adiabatic...