21 HEAT, WORK, AND THE FIRST LAW OF THERMODYNAMICS

[Pages:2]CHAPTER 21 HEAT, WORK, AND THE FIRST LAW OF THERMODYNAMICS

ActivPhysics can help with these problems: Activities 8.5-8.13

Section 21-1: The First Law of Thermodynamics

Problem

1. In a perfectly insulated container, 1.0 kg of water is stirred vigorously until its temperature rises by 7.0?C. How much work was done on the water?

Solution

Since the container is perfectly insulated thermally, no heat enters or leaves the water in it. Thus, Q = 0 in Equation 21-1. The change in the internal energy of the water is determined from its temperature rise,

AU = rnc AT (see comments in Section 19-4 on internal energy), so W = -AW = -(1 kg) x

(4.184 kJ/kg.K)(7 K) = -29.3 kJ. (The negative sign signifies that work was done on the water.)

-.

Section 21-2: Thermodynamic Processes .

Problem

8. An ideal gas expands from the state (PI,Vl) to the state ( 9 , Vz), where P2= 2P1 and V2 = 2V1. The expansion proceeds along the straight diagonal path AB shown in Fig. 21-26. Find an expression for the work done by the gas during this process.

Problem

5. The most efficient large-scale electric power generating systems use high-temperature gas turbines and a so-called combined cycle system that maximizes the conversion of thermal energy into useful work. One such plant produces electrical energy at the rate of 360 hlW, while extracting energy-from its natural gas fuel at the rate of 670 MW. (a) At what rate does it reject waste heat to the environment? (b) Find its efficiency, defined as the percent of the total energy extracted from the fuel that ends up as work.

Solution

(a) If we assume that the generating system operates in a cycle and choose it as "the system," then dli/dt = 0 and Equation 21-2 implies dQ/dt = dW/dt. Here, dW/dt is the rate that the generator supplies energy to its surroundings (360 h/lW in this problem) and dQ/dt is the net rate of heat flow into the generator from the surroundings. Since the system is just the generator, the net heat flow is the difference between the heat extracted from its fuel and the heat exhausted to the

environment, i.e., dQ/dt = (dQ/dt)i, - (dQ/dt)our = - -

670 MW - (dQ/dt),,,. Therefore, (dQ/dt),,, = 670 MW - 360 MW = 310 MW. (Note: If the system

is assumed to be the generator and its fuel, as in Example 21-1, then dW/dt is still 360 MW, but the system's internal energy decreases because energy is

+ extracted from the fuel, dU/dt = -670 MW, and there

is no heat input. Then dQ/dt = -670 MW 360 MW = -310 MW, representing the rate of heat rejected to the environment.) (b) The efficiency is (dW/dt)/(dQ/dt)i, = 360 MW/670 MW = 53.7% (see Section 22-2).

FIGURE 21-26 Problems 8 and 9.

Solution

The work done by the gas equals the area under the straight diagonal path AB in Fig. 21-26. The area of

+ + this trapezoid is W = +(PI P2)(V2- Vl ) = +(PI 2 2Pl)( 2 4 - Vl) = PIv]. W can also be obtained from

+ Equation 21-3. On the path AB, P = PI (V - Vl)x

(P2 - PI)/(% - Vl). Then

Problem

. .

11. A balloon contains 0.30 mot of helium. It rises, while maintaining a constant 300 K temperature, to an altitude where its volume has expanded 5 times. How much work is done by the gas in the balloon during this isothermal expansion? Neglect tension forces in the balloon.

Solution

During an isothermal expansion, the work done by a given amount of ideal gas is W = nR'TIn(h/V1) =

(0.3 mo1)(8.314 J/mol.K)(300 K)ln(5) = 1.20 kJ (see

Equation 21-4).

Problem

19. A gas with y = 1.4 is at 100 kPa pressure and

occupies 5.00 L. (a) How much work does it take

to compress the gas adiabatically to 2.50 L?

.

(b) What is its final pressure?

Solution

The work done by an ideal gas undergoing an

adiabatic process is W12= ( P I K - P2V2)/(y - 1) (see

Equation 21-14). Since the compression is specified by given values of PI, Vl, and V2, we first find the final pressure from the adiabatic gas law. (b) P2 = Pl (Vl+ V2l7 = (100 kPa)(5 L12.5 L)1.4= 264 kPa. (a) Then the work done on the gas (which is -W12) is -W12 = (P2V2-PlVl)/(y-l) = [(264 kPa)(2.5 L)-(100 kPa)x (5 L)]/0.4 = 399 J.

25:By how much must the volume of a gas with 7 = 1.4 be changed in an adiabatic process if the kelvin temperature is to double?

Solution

V/G = (T~/T)'/(Y-') = ( 0 . 5 ) ' / ~ =. ~0.177

(Equation 21-13b).

Problem 31. An ideal gas with y = 1.67 starts a t point A in

Fig. 21-29, where its volume and pressure are 1.00 m3 and 250 kPa, respectively. It then undergoes an adiabatic expansion that triples its volume, ending at point B. It's then heated a t constant volume to point C, then compressed isothermally back to A. Find (a) the pressure a t B, (b) the pressure at C, and (c) the net work done on the gas.

FIGURE 21-29 Problem 31.

Solution

(a) From the adiabatic law for an ideal gas (Equation 21-13a), PB= PA(v~/VB)7= (250 kPa) x

= 39.9 kPa. (b) Point C lies on an isotherm with A, so the ideal gas law (Equation 20-2) yields

+ + PC= PAVA/VC= (250 kPa)(i) = 83.3 kPa.

(c) Wnet= WAB WBC WCA.WABis for an adiabatic process (Equation 21-14) and equals

(PAVA - P B V B ) / ( ~- 1) = [(250 kPa)(1 m3) (39.9 kPa)(3 m3)]/0.67 = 194 kJ; WBC is for an

isovolumic process and equals zero; WCA is for an isothermal Process (Equation 21-4) and equals

~ WI nA( v ~ / v c )= PAVAl n ( v ~ / V =~ )250 kJ In(?) = -275 kJ. Thus, Wnet = -80.2 kJ. The work done on

the gas is the negative of this.

Problem 37. A bicycle pump consists of a cylinder 30 cm long when the pump handle is all the way out. The pump contains air (7 = 1.4) at 20?C. If the pump outlet is blocked and the handle pushed until the internal length of the pump cylinder is 17 cm, by how much does the air temperature rise? Assume that no heat is lost.

.

Solution If no heat is lost (or gained) by the gas, the compression is adiabatic and Equation 21-13b gives TV7-' = ~ ~ ~ o 7Th- elre. fore, the temperature rise is

T - To = A T = To[(Vo/V)7-I - 1). Since Vo/V = (30 cm/17 cm), AT = [(30/17)O.~- 1](293K) =

74.7 CO.

Problem 43. A mixture of monatomic and diatomic gases has

specific heat ratio y = 1.52. What fraction of the molecules are monatomic?

Solution

+ The internal energy of a mixture of two ideal gases is

U = flN E ~ fiNE2, where fi is the fraction of the total number of molecules, N, of type 1, and El is

the average energy of a molecule of type 1, etc.

Classically, E = g(;kT), where g is the number of

degrees of freedom. The molar specificheat at constant volume is Cv = ( i ) ( d ~ / d T =) ( N ~ / N ) d l d T x

(flNgl;kT + f2Ng24kT) = @(fig1 + f2g2). Suppose

that the temperature range is such that 91 = 3 for the

+ monatomic gas, and 92 = 5 for the diatomic gas, as

discussed in Section 21-3. Then Cv = R(1.5fl 2.5f2) = R(2.5 - fl), where f2 = 1- fl since the sum

+ of the fractions of the mixture is one. Now, Cv can

also be specified by the ratio 7 = Cp/Cv = 1 R/CV,

or Cv = R/(y - I), so in this problem, 2.5 - f l =

110.52,o r 5 = 57.7%.

Problem 45. A gas mixture contains monatomic argon and

diatomic oxygen. An adiabatic expansion that doubles its volume results in the pressure dropping to one-third of its original value. What fraction of the molecules are argon?

Solution From the pressures and volumes in the described adiabatic expansion, PoV2 = (iPo)(2Vo)7,we can calculate that y = In 3/ In 2 = 1.58. Then the result of

Problem 43 gives 2.5 - fA, = 110.58, Or fAr = 79.0%.

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