Chapter 3 Solution Manual

Thermodynamics 1 Hipolito B. Sta. Maria Chapter 3 : The Ideal Gas Solution Manual

1. An automobile tire is inflated to 32psig pressure at 50 F. After being driven, the temperature rises to 75 F. Determine the final gage pressure assuming the volume remains constant. Ans. 34.29 psig (Electrical Engineering Board Exam Problem) Given: P1 = 32 psig T1 = 50 F + 460 = 510 R T2 = 75 F + 460 = 535 R

Required: P2 = ?

Solution: Since volume is constant, use Charles’ Law on constant volume processes. 𝑃1 𝑃2 = 𝑇1 𝑇2 Note: P and T should be absolute. 𝑃1 𝑇2 𝑇1 (32 𝑝𝑠𝑖𝑔 + 14.696 𝑝𝑠𝑖)( 535 R ) 𝑃2 = 510 R 𝑃2 = 48.9850 𝑃2 =

Converting back to gage pressure 𝑃2 = 48.9850 − 14.696 𝑃2 = 34.29 𝑝𝑠𝑖𝑔 2. If 100 ft3 of atmospheric air at zero Fahrenheit temperature are compressed to a volume of 1 ft3 at a temperature of 200 F, what will be the pressure of the air in psi? Ans. 2109 psia (Electrical Engineering Board Exam Problem) Given: V1 = 100 ft3 T1 = 0 F + 460 = 460 R P1 = 14.696 psia

V2 = 1 ft3 T2 = 200 F + 460 = 660 R

Solution: 𝑃1 𝑉1 𝑃2 𝑉2 = 𝑇1 𝑇2 𝑃1 𝑉1 𝑇2 𝑃2 = 𝑇1 𝑉2 (14.696 psia )(100 ft 3 )(660 R) 𝑃2 = (460 R)(1 ft 3 ) 𝑃2 = 2109 𝑝𝑠𝑖𝑎

Required: P2 = ?

3. A 10 ft3 tank contains gas at a pressure of 500 psia, temperature of 85 F and a weight of 25 lbs. A part of the gas was discharged and the temperature and pressure changed to 70 F and 300 psia, respectively. Heat was applied and the temperature was back to 85 F. Find the final weight, volume, and pressure of the gas. Ans: 15.43 lb; 10 ft3; 308.5 psia (Electrical Engineering Board Exam Problem) Given: V1 = 10 ft3 P1 = 500 psia T1 = 85 F + 460 = 545 R m = 25 lb

Required: m=? V=? P3 = ?

T2 = 70 F + 460 = 330 R P2 = 300 psia T3 = 85 F + 460 = 545 R

Solution: The volume is fixed – it remains 10 ft3 throughout. V1 = 10 ft3 = V2 𝑃𝑉 = 𝑚𝑅𝑇 𝑃1 𝑉1 𝑃2 𝑉2 = 𝑚1 𝑅𝑇1 𝑚2 𝑅𝑇2 𝑃2 𝑉2 𝑚1 𝑇1 𝑚2 = 𝑃1 𝑉1 𝑇2 (300 psia)(10 ft 3 )(25 lb)(545 R) 𝑚2 = (500 psia)(10 ft 3 )(530 R) 𝑚2 = 15.42 𝑙𝑏

𝑃2 𝑃3 = 𝑇2 𝑇3 𝑃2 𝑇3 𝑃3 = 𝑇2 (300 psia)(545 R) 𝑃3 = 530 R 𝑃2 = 308.49 𝑝𝑠𝑖𝑎

4. Four hundred cubic centimeters of gas at a pressure of 740 mmHg absolute and temperature of 18 C undergoes a process until the pressure and temperature becomes 760 mmHg absolute and 0 C respectively. Calculate the final volume of the gas. Ans. 365 cc (Electrical Engineering Board Exam Problem) Given: V1 = 400 cm3 P1 = 740 mmHg = 74 cmHg T1 = 18 C + 273 = 291 K Solution: 𝑃1 𝑉1 𝑃2 𝑉2 = 𝑇1 𝑇2 𝑃1 𝑉1 𝑇2 𝑉2 = 𝑇1 𝑃2 (74 cmHg)(400 cm3 )(273 K) 𝑉2 = (291 K)(76 cmHg) 𝑉2 = 365 𝑐𝑐

Required: V=? P2 = 760 mmHg = 76 cmHg T2 = 0 C + 273 = 273 K

5. A motorist equips his automobile tires with a relief-type valve so that the pressure inside the tire will never will exceed 240 kPa (gage). He starts a trip with a pressure of 200 kPa (gage) and a temperature of 23 C in the tires. During the long drive, the temperature of the air in the tires reaches 83 C. Each tire contains 0.11 kg of air. Determine (a) the mass of air escaping each tire, (b) the pressure of the tire when the temperature returns to 23 degree Celsius. Ans (a) 0.0064 kg; (b) 182.48 kPa (gage) 𝑁∙𝑚

Given:

𝑅 = 287.08 𝑘𝑔 ∙𝐾

Required:

m = 0.11 kg P1 = 200 kPa

T1 = 23C + 273 = 296 K T2 = 83 C + 273 = 356 K

m=? P=?

Solution: 𝑃𝑉 = 𝑚𝑅𝑇 𝑚𝑅𝑇 𝑉= 𝑃 𝑁 ∙ 𝑚 (0.11 kg)(287.08 )(296 K) 𝑘𝑔 ∙ 𝐾 𝑉= 200000 Pa + 101325 Pa 𝑉 = 0.0310 𝑚3 (240000 𝑃𝑎 + 101325 𝑃𝑎)(0.0310 𝑚3 ) 𝑃2 𝑉2 = 𝑁 ∙ 𝑚 𝑅 𝑇2 (287.08 )(356 K) 𝑘𝑔 ∙ 𝐾 𝑚2 = 0.1036 𝑘𝑔 𝑚2 =

𝑚 − 𝑚2 = 0.11 𝑘𝑔 − 0.1036 𝑘𝑔 𝑚 = 0.0064 𝑘𝑔 𝑃𝑉 = 𝑚𝑅𝑇 𝑚𝑅𝑇 𝑃= 𝑉 𝑁 ∙ 𝑚 (0.1036 𝑘𝑔)(287.08 )(296 K) 𝑘𝑔 ∙ 𝐾 𝑃= (0.0310 𝑚3 ) 𝑃 = 283.80 𝑘𝑃𝑎 − 101.325 𝑘𝑃𝑎 𝑃 = 183 𝑘𝑃𝑎

6. A 6-m3 tank contains helium at 400K and is evacuated from atmospheric pressure to a pressure of 740 mm Hg vacuum. Determine (a) mass of helium remaining in the tank, (b) mass of helium pump out, (c) the temperature of the helium falls to 10 C. What is the pressure in kPa? Ans. (a) 0.01925 kg; (b) 0.7123 kg; (c) 1.886 kPa Given: V1 = 6-m3 T1 = 400 K

T2 = 10 C + 273 = 283 K 𝑁∙𝑚 𝑅 = 2077.67 𝑘𝑔 ∙𝐾

Required: a. m = ? b. m = ?

Pg = 740 mmHg vacuum

Patm = 760 mmHg

c. P = ? kPa

Solution: 𝑃𝑎𝑏𝑠 = 𝑃𝑎𝑡𝑚 − 𝑃𝑔 b. 𝑚2 = 𝑚 − 𝑚1 = 0.7315𝑘𝑔 − 0.01925𝑘𝑔 𝑚2 = 0.71225 𝑘𝑔 𝑃𝑎𝑏𝑠 = 760 mmHg − 740 mmHg = 20mmHg = 2.66 𝑘𝑃𝑎 c. 𝑃𝑉 = 𝑚𝑅𝑇 (101.325 𝑘𝑃𝑎)(6m3 ) 𝑃1 𝑉1 𝑚= = = 0.7315 𝑘𝑔 𝑘𝑁 ∙ 𝑚 𝑘𝑁 ∙ 𝑚 𝑅 𝑇1 (0.01925 𝑘𝑔) (2.07767 ) (283 𝐾) (2.07767 ) (400 K) 𝑚𝑅T 𝑘𝑔 ∙ 𝐾 𝑘𝑔 ∙ 𝐾 𝑃= = (6m3 ) 𝑉 a. 𝑃𝑉 = 𝑚𝑅𝑇 3 𝑃 = 1.886 𝑘𝑃𝑎 (2.666 𝑘𝑃𝑎)(6m ) 𝑃1 𝑉1 𝑚1 = = 𝑘𝑁 ∙ 𝑚 𝑅 𝑇1 (2.07767 ) (400 K) 𝑘𝑔 ∙ 𝐾 𝑚1 = 0.01925 𝑘𝑔

7. An automobile tire contains 3730 cu in. of air at 32 psig and 80 F. (a) What mass of air is in the tire? (b) In operation, the air temperature increases to 145 F. If the tire is inflexible, what is the resulting percentage increase in gage pressure? (c) What mass of the 145 F air must be bled off to reduce the pressure back to its original value? Ans. (a) 0.5041 Ib; (b) 17.53%; (c) 0.0542 lb Given: V1 = 3730 in3 P 1 = 32 psig

Required: a. m = ? b. %P = ?

T2 = 145 F + 460 = 605 R 𝑙𝑏 ∙ 𝑓𝑡 𝑅 = 53.342 𝑙𝑏 ∙𝑅

T1 = 80 F + 460 = 540 R

c. m = ?

Solution: 37.63 𝑝𝑠𝑖𝑔 − 32 𝑝𝑠𝑖𝑔 a. 𝑃𝑉 = 𝑚𝑅𝑇 %𝑃 = ×100 3 (32 𝑝𝑠𝑖𝑔 + 14.692 𝑝𝑠𝑖𝑎)(3730 in ) 𝑃𝑉 32 𝑝𝑠𝑖𝑔 𝑚1 = = %𝑃 = 17.59 % 𝑙𝑏 ∙ 𝑓𝑡 𝑖𝑛 𝑅 𝑇1 (53.342 ) (540 R)(12 ) 𝑙𝑏 ∙ 𝑅 𝑓𝑡 𝑚1 = 0.50386 𝑙𝑏 c. 𝑃𝑉 = 𝑚𝑅𝑇 (32 𝑝𝑠𝑖𝑔 + 14.692 𝑝𝑠𝑖𝑎)(3730 in3 ) 𝑃𝑉 𝑚 = = 2 b. V1 = 3730 in3 = V2 𝑙𝑏 ∙ 𝑓𝑡 𝑖𝑛 𝑅 𝑇1 (53.342 ) (605 R)(12 ) 𝑃1 𝑃2 𝑙𝑏 ∙ 𝑅 𝑓𝑡 = 𝑚2 = 0.45232 𝑙𝑏 𝑇1 𝑇2 (32 𝑝𝑠𝑖𝑔 + 14.692 𝑝𝑠𝑖𝑎)(605 R) 𝑃2 = 𝑚3 = 𝑚1 − 𝑚2 = 0.50386 𝑙𝑏 − 0.45232 𝑙𝑏 540 R 𝑚3 = 0.05154 𝑙𝑏 𝑃2 = 52.6148 𝑝𝑠𝑖𝑎 − 14.692 𝑝𝑠𝑖𝑎 = 37.63 𝑝𝑠𝑖𝑔 8. A spherical balloon is 40 ft in diameter and surrounded by air at 60 F and 29.92in Hg abs. (a) If the balloon is filled with hydrogen at a temperature of 70 F and atmospheric pressure, what total load can it lift? (b) If it contains helium instead of hydrogen, other conditions remaining the same, what load can it lift? (c) Helium is nearly twice as heavy as hydrogen. Does it have half the lifting force? R for hydrogen is 766.54 and for helium is 386.04 ft.lb/lb. R. Ans. (a) 2381 lb; (b) 2209 lb Given: d = 40 ft T = 60 F + 460 = 520 R

Thydrogen = 70 F + 460 = 530 R 𝑙𝑏 ∙ 𝑓𝑡 𝑅hydrogen = 766.54 𝑙𝑏 ∙𝑅

Required: a. m = ? b. m = ?

P = 29.92 in Hg abs

𝑅helium = 386.04

c.

Solution: 4 𝑉 = 3 𝜋(20𝑓𝑡)3 = 33510.32 𝑓𝑡 3 𝑃𝑉 = 𝑚𝑅𝑇 𝑚𝑎𝑖𝑟 =

𝑃𝑉 𝑅𝑇

=

14.692 𝑝𝑠𝑖𝑎 144𝑖𝑛2 (29.92 in Hg × )(33510.32 𝑓𝑡 3 )( ) 29.92 in Hg 𝑓𝑡2

(53.342

𝑙𝑏 ∙ 𝑓𝑡 )(520 R) 𝑙𝑏 ∙𝑅

𝑙𝑏 ∙ 𝑓𝑡 𝑙𝑏 ∙𝑅

𝑚𝑙𝑖𝑓𝑡 = 𝑚𝑎𝑖𝑟 − 𝑚ℎ𝑦𝑑𝑟𝑜𝑔𝑒𝑛 = 2557.4201 𝑙𝑏 − 174.6014 𝑙𝑏

𝑚𝑙𝑖𝑓𝑡 = 2382.8187 𝑙𝑏 a. b. 𝑃𝑉 = 𝑚𝑅𝑇 𝑃𝑉 𝑅𝑇

144𝑖𝑛2 ) 𝑓𝑡2

(14.692 𝑝𝑠𝑖𝑎)(33510.32 𝑓𝑡 3 )(

𝑚𝑎𝑖𝑟 = 2557.4201 𝑙𝑏

𝑚ℎ𝑒𝑙𝑖𝑢𝑚 =

a. 𝑃𝑉 = 𝑚𝑅𝑇

𝑚ℎ𝑒𝑙𝑖𝑢𝑚 = 346.6971 𝑙𝑏 𝑚𝑙𝑖𝑓𝑡 = 𝑚𝑎𝑖𝑟 − 𝑚ℎ𝑒𝑙𝑖𝑢𝑚 = 2557.4201 𝑙𝑏 − 346.6971 𝑙𝑏 𝑚𝑙𝑖𝑓𝑡 = 2210.723 𝑙𝑏 c. No

𝑚ℎ𝑦𝑑𝑟𝑜𝑔𝑒𝑛 =

𝑃𝑉 𝑅𝑇

=

144𝑖𝑛2 (14.692 𝑝𝑠𝑖𝑎)(33510.32 𝑓𝑡 3 )( ) 𝑓𝑡2 𝑙𝑏 ∙ 𝑓𝑡 (766.54 )(530 R) 𝑙𝑏 ∙𝑅

𝑚ℎ𝑦𝑑𝑟𝑜𝑔𝑒𝑛 = 174.6014 𝑙𝑏

=

(386.04

𝑙𝑏 ∙ 𝑓𝑡 )(530 R) 𝑙𝑏 ∙𝑅

9. A reservoir contains 2.83 cu m of carbon monoxide at 6895 kPa and 23.6 C. An evacuated tank is filled from the reservoir to a pressure of 3497 kPa and a temperature of 12.4 C, while the pressure in the reservoir decreases to 6205 kPa and the temperature to 18.3 C. What is the volume of the tank? R for CO is 296.92 J/kg.K. Ans. 0.451 m Given: Vco = 2.83 m3

𝑅co = 296.92

P1 = 6895 kPa T1 = 23.6 C + 273 = 296.6 K

P2 = 3497 kPa T2 = 12.4 C + 273 = 285.4 K

Solution: 𝑃𝑉 = 𝑚𝑅𝑇 (6895 kPa)(2.83m3 ) 𝑃1 𝑉𝑐𝑜 𝑚1 = = 𝑘𝑁 ∙ 𝑚 𝑅 𝑇1 (0.29692 ) (296.6 K) 𝑘𝑔 ∙ 𝐾 𝑚1 = 221.5696 𝑘𝑔 𝑚𝑡𝑎𝑛𝑘 =

Required: Vtank = ?

𝐽 𝑘𝑔 ∙𝐾

Ptank = 6205 kPa Ttank = 18.3 C +273 = 291.3 K

𝑚𝑡𝑎𝑛𝑘 = 203.024 𝑘𝑔 𝑚 = 𝑚1 − 𝑚𝑡𝑎𝑛𝑘 = 221.5696 𝑘𝑔 − 203.024 𝑘𝑔 𝑚 = 18.5451 𝑘𝑔 𝑃𝑉 = 𝑚𝑅𝑇 3)

(6205 kPa)(2.83m 𝑃𝑡𝑎𝑛𝑘 𝑉𝑐𝑜 = 𝑘𝑁 ∙ 𝑚 𝑅 𝑇𝑡𝑎𝑛𝑘 (0.29692 ) (291.3 K) 𝑘𝑔 ∙ 𝐾

𝑉𝑡𝑎𝑛𝑘

𝑁 ∙ 𝑚 𝑚𝑅𝑇2 (18.5451 𝑘𝑔) (296.92 𝑘𝑔 ∙ 𝐾 ) (285.4 K) = = (3497 kPa) 𝑃2

𝑉𝑡𝑎𝑛𝑘 = 0.449 𝑚3

10. A gas initially at 15 psia and 2 cu ft undergoes a process to 90 psia and 0.60 cu ft, during which the enthalpy increases by 15.5 Btu; cv = 2.44Btu/lb. R. Determine (a) ΔU, (b) cp, and (c) R. Ans. (a) 11.06 Btu; (b) 3.42 Btu/lb.R; (c) 762.4ft.lb/lbR Given: P1 = 15 psia V1 = 2 ft3

P2 = 90 psia V2 = 0.60 ft3

ΔH = 15.5 BTU 𝐵𝑇𝑈 𝑐𝑣 = 2.44 𝑙𝑏 ∙ °𝑅

Required: a. ΔU = ? b. Cp = ? c. R

Solution: ∆𝑊𝑓 = {(90psia×0.60 ft 3 ) − (15psia×2 ft 3 ) }× ∆𝑊𝑓 = 4.0720 𝐵𝑇𝑈

a. ΔH = 15.5 Btu = Qp 𝑄𝑝 = ∆𝑈 + ∆𝑊𝑓 ∆𝑈 = 𝑄𝑝 − ∆𝑊𝑓 ∆𝑈 = 15.5 𝐵𝑇𝑈 − 4.0720 𝐵𝑇𝑈 ∆𝑈 = 11.06 𝐵𝑇𝑈 b. ∆U = 11.06 BTU = Qv 𝑄𝑣 = 𝑚𝐶𝑣 (𝑇2 − 𝑇1 ) 𝑄𝑣 11.06 𝐵𝑇𝑈 𝑚(𝑇2 − 𝑇1 ) = = 𝐶𝑣 2.44 𝐵𝑇𝑈 𝑙𝑏 ∙ °𝑅 𝑚(𝑇2 − 𝑇1 ) = 4.53 𝑙𝑏 ∙ °𝑅

144 𝑖𝑛2 𝐵𝑇𝑈 778 𝑓𝑡 3 ∙ 𝑙𝑏

𝑄𝑝 = 𝑚𝐶𝑝 (𝑇2 − 𝑇1 ) 𝑄𝑝 𝐶𝑝 = 𝑚(𝑇2 − 𝑇1 ) 15.5 BTU 𝐶𝑝 = 4.53 𝑙𝑏 ∙ °𝑅 𝐵𝑇𝑈 𝐶𝑝 = 3.42 𝑙𝑏 ∙ °𝑅 c. 𝐶𝑝 = 𝐶𝑣 + 𝑅 𝑅 = 𝐶𝑝 − 𝐶𝑣 𝐵𝑇𝑈 𝐵𝑇𝑈 𝑅 = 3.42 − 2.44 𝑙𝑏 ∙ °𝑅 𝑙𝑏 ∙ °𝑅 𝐵𝑇𝑈 𝑓𝑡 ∙ 𝑙𝑏 𝑅 = 0.98 × 778 𝑙𝑏 ∙ °𝑅 𝐵𝑇𝑈 𝑓𝑡 ∙ 𝑙𝑏 𝑅 = 762.44 𝑏 ∙ °𝑅

11. For a certain gas, R = 0.277 kJ/kg.K and k = 1.384. (a) What are the value of cp and cv? (b) What mass of this gas would occupy a volume of 0.425 cu m at 517.11 kPa and 26.7 C? (c) If 31.65 kJ are transferred to this gas at constant volume in (b), what are the resulting temperature and pressure? Ans. (a) 0.7214 and 0.994 kJ/kg.R; (b)2.647 kg; (c) 43.27 C, 545.75 kPa Given: 𝑘𝐽 𝑅 = 0.277 𝑘𝑔∙𝐾

Required: a. cp = ?, cv = ?

𝑉 = 0.425 𝑚

k = 1.384 Qv = 31.65 kJ

P = 517.11 kPa T = 26.7 C + 273 = 299.7 K

3

b. m = ? c. T2 = ?, P2 = ?

Solution:

a. 𝐶𝑝 =

𝑘𝑅 𝑘−1

=

𝑘𝐽 ) 𝑘𝑔∙𝐾

(1.384)( 0.277

𝑘𝐽 𝐶𝑝 = 0.998 𝑘𝑔 ∙ 𝐾

1.384 − 1

𝑘𝐽 ( 0.277 ) 𝑅 𝑘𝑔 ∙ 𝐾 𝐶𝑣 = = 𝑘−1 1.384 − 1 𝑘𝐽 𝐶𝑣 = 0.7214 𝑘𝑔 ∙ 𝐾 b. 𝑃𝑉 = 𝑚𝑅𝑇 (517.11 kPa)(0.425 𝑚 3 ) 𝑃𝑉 𝑚= = 𝑘𝐽 𝑅𝑇 (0.277 ) (299.7 K) 𝑘𝑔 ∙ 𝐾 𝑚 = 2.647 𝑘𝑔

c. 𝑄𝑣 = 𝑚𝐶𝑣 (𝑇2 − 𝑇1 ) 𝑄𝑣 𝑇2 = + 𝑇1 𝑚𝐶𝑣 31.65 𝑘𝐽 𝑇2 = (2.647 𝑘𝑔) (0.7214

𝑘𝐽 ) 𝑘𝑔 ∙ 𝐾

+ 299.7 K

𝑇2 = 316.28 𝐾 𝑃1 𝑃2 = 𝑇1 𝑇2 𝑃1 𝑇2 (517.11 kPa)(316.28 𝐾) 𝑃2 = = 𝑇1 299.7 K 𝑃2 = 545.72 𝑘𝑃𝑎