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Home , Heat, Isobaric process, Isothermal process, Solution, Volume (thermodynamics)

to HW#14 CJ5 15.CQ.006. REASONING AND SOLUTION The text drawing shows a - graph in which a undergoes a two-step process from A to B and from B to C.

From A to B: The volume V of the gas increases at constant pressure P. According to the law (Equation 14.1), PV = nRT , the T of the gas must increase. According to Equation 14.7, U = (3 / 2)nRT , if T increases, then ∆U , the change in the internal , must be positive. Since the volume increases at constant pressure (∆V increases), we know from Equation 15.2, W = P∆V , that the done is positive. The first law of (Equation 15.1) states that ∆U = Q − W ; since ∆U and W are both positive, Q must also be positive.

From B to C The pressure P of the gas increases at constant volume V. According to the (Equation 14.1), PV = nRT , the temperature T of the gas must increase. According to Equation 14.7, U = (3 / 2)nRT , if T increases, then ∆U , the change in the , must be positive. Since the process occurs isochorically (∆V = 0 ), and according to Equation 15.2, W = P∆V, the work done is zero. The first law of thermodynamics (Equation 15.1) states that ∆U = Q − W ; since W = 0, Q is also positive since ∆U is positive.

These results are summarized in the table below:

∆U Q W

A → B + + +

B → C + + 0

CJ5 15.CQ.007. REASONING AND SOLUTION Since the process is an , Q = 0. Since the gas expands into chamber B under zero external pressure, the work done by the gas is W = P∆V = 0. According to the first law of thermodynamics, the change in the internal energy is, therefore, zero: ∆U = Q − W = 0 . The internal energy of an ideal gas is proportional to the temperature of the gas (Equation 14.7). Since the change in the internal energy of the gas is zero, the temperature change of the gas is zero. The final temperature of the gas is the same as the initial temperature of the gas. CJ5 15.CQ.012. REASONING AND SOLUTION According to the second law of thermodynamics, flows spontaneously from a substance at a higher temperature to a substance at a lower temperature and does not flow spontaneously in the reverse direction. Therefore, according to the second law of thermodynamics, work must be done to remove heat from a substance at a lower temperature and deposit it in a substance at a higher temperature. In other words, the second law requires that energy in the form of work must be supplied to an air conditioner in order for it to remove heat from a cool space and deposit the heat in a warm space. An advertisement for an automobile that claimed the same gas mileage with and without the air conditioner would be suspect. Since the car would use more energy with the air conditioner on, the car would use more gasoline. Therefore, the mileage should be less with the air conditioner running.

CJ5 15.P.001. SSM REASONING Since the change in the internal energy and the heat released in the process are given, the first law of thermodynamics (Equation 15.1) can be used to find the work done. Since we are told how much work is required to make the car go one mile, we can determine how far the car can travel. When the gasoline burns, its internal energy decreases and heat flows into the surroundings; therefore, both ∆U and Q are negative.

SOLUTION According to the first law of thermodynamics, the work that is done when one gallon of gasoline is burned in the is

WQ=−∆U=−10. 0×1088 J – (–1.19 ×10 J) = 0.19 ×108 J

Since 6.0 ×105 J of work is required to make the car go one mile, the car can travel

8 F 1 mile I 01. 9×10 J G J = 32 miles H 6.0 ×105 J K CJ5 15.P.007. REASONING The work done during an is the product − of the pressure P and the change in volume, Vf Vi, where Vf and Vi are the final and initial , respectively (see Equation 15.2).

SOLUTION

a. The work done when the volume is compressed is

=−=×53×−−33−×3=−×2 WPcVfiVh c1..5 10 Pahc2 0 10 m 7.0 10 m h 7.5 10 J

b. The work done when the volume expands is

=−=×53×−−33−×3=×2 WPcVfiVh c1..5 10 Pahc8 0 10 m 2.0 10 m h 9.0 10 J CJ5 15.P.011. REASONING AND SOLUTION The work done by the expanding gas is

W = Q − ∆U = 2050 J − 1730 J = 320 J

The work, according to Equation 6.1, is also the magnitude F of the exerted on the piston times the magnitude s of its displacement. But the force is equal to the mg of the block and piston, so that the work is W = Fs = mgs. Thus, we have

W 320 J s == =0.24 m mg bg135 kg c9.80 m / s2 h CJ5 15.P.018. REASONING AND SOLUTION For an Q = W = nRT ln (Vf/Vi), where Vf/Vi = 4.00. Thus,

W =⋅(.3 00 mol)8.31J /bmol Kg (373 K)ln(4.00)=1.29 ×104 J

CJ5 15.P.020. REASONING AND SOLUTION As Section 14.1 discusses, the number of moles n is given by the m divided by the mass per :

m 60. g n ===15. mol Mass per mole 4.0 g / mol

For an isothermal process we have (see Equation 15.3)

F V I W 9600 J ln f == = 2.08 G J ⋅ H Vi K nRT bg1.5 mol 8.31J / bmol Kgb370 Kg

2.08 Therefore, Vf/Vi = e =8.0. CJ5 15.P.034. REASONING AND SOLUTION a. The amount of heat needed to raise the temperature of the gas at constant volume ∆ ∆ is given by Equations 15.6 and 15.8, Q = n CV T. Solving for T yields

Q 5.24 × 103 J ∆T == =×1.40 102 K 3 nCv b3.00 molgb 2 Rg

b. The change in the internal energy of the gas is given by the first law of thermodynamics with W = 0, since the gas is heated at constant volume:

∆UQ=−W=5..24 ×1033J0−=5 24 ×10 J

c. The change in pressure can be obtained from the ideal gas law,

nR∆T bg3..00 mol Rc1 40 × 102 Kh ∆P == =×23. 3 103 Pa V 1.50 m 3 CJ5 15.P.043. REASONING AND SOLUTION a. The efficiency is e = W/QH, so that

QH = W/e = (5500 J)/(0.64) = 8600 J

b. The rejected heat is − − QC = QH W = 8600 J 5500 J = 3100 J CJ5 15.P.046. REASONING AND SOLUTION The maximum efficiency is given by

− − emax = 1 (TC/TH) = 1 [(620 K)/(950 K)] = 0.35

Since the engine operates at three-fifths maximum efficiency, its efficiency is

e = 0.60 emax = (0.60)(0.35) = 0.21

CJ5 15.P.061. REASONING AND SOLUTION a. The work is − − W = QH QC = 3140 J 2090 J = 1050 J

b. The coefficient of performance (CP) is

CP = QH/W = (3140 J)/(1050 J) = 2.99