CHAPTER 31 IDEAL GAS LAWS EXERCISE 144, Page 317 1. The pressure of a mass of gas is increased from 150 kPa to 750 kPa at constant temperature. Determine the final volume of the gas, if its initial volume is 1.5 m 3 . Since the change occurs at constant temperature (i.e. an isothermal change), Boyle's law applies, 3 i.e. p 1 V 1 = p 2 V 2 where p 1 = 150 kPa, p 2 = 750 kPa and V 1 = 1.5 m Hence, (150)(1.5) = (750)V 2 150 1.5 from which, volume V = = 0.3 m 3 2 750 2. In an isothermal process, a mass of gas has its volume reduced from 50 cm 3 to 32 cm 3 . If the initial pressure of the gas is 80 kPa, determine its final pressure. Since the change occurs at constant temperature (i.e. an isothermal change), Boyle's law applies, 3 3 i.e. p 1 V 1 = p 2 V 2 where p 1 = 80 kPa, V 1 = 50 cm and V 2 = 32 cm Hence, (80)(50) = (p 2 )(32) 80 50 from which, pressure p = = 125 kPa 2 32 1 3. The piston of an air compressor compresses air to of its original volume during its stroke. 4 Determine the final pressure of the air if the original pressure is 100 kPa, assuming an isothermal change. Since the change occurs at constant temperature (i.e. an isothermal change), Boyle's law applies, 354 © John Bird Published by Taylor and Francis 1 i.e. p V = p V where p = 100 kPa, V = V 1 1 2 2 1 2 4 1 1 Hence, (100)( V ) = (p )( V ) 1 2 4 1 100 V 100 from which, pressure p = 1 = 400 kPa 2 11 V 441 4. A quantity of gas in a cylinder occupies a volume of 2 m 3 at a pressure of 300 kPa. A piston slides in the cylinder and compresses the gas, according to Boyle's law, until the volume is 0.5 m 3 . If the area of the piston is 0.02 m 2 , calculate the force on the piston when the gas is compressed. An isothermal process means constant temperature and thus Boyle's law applies, i.e. p 1 V 1 = p 2 V 2 3 3 where V 1 = 2 m , V 2 = 0.5 m and p 1 = 300 kPa. Hence, (300)(2) = p 2 (0.5) 300 2 from which, pressure, p = = 1200 kPa 2 0.5 force Pressure = , from which, force = pressure area. area Hence, force on the piston = (1200 10 3 Pa)(0.02 m 2 ) = 24000 N = 24 kN 5. The gas in a simple pump has a pressure of 400 mm of mercury (Hg) and a volume of 10 mL. If the pump is compressed to a volume of 2 mL, calculate the pressure of the gas, assuming that its temperature does not change? Since PV11 PV 2 2 then (400 mm Hg)(10 mL) = ( P2 )(2 mL) 355 © John Bird Published by Taylor and Francis 400mm Hg 10mL from which, new pressure, P = = 2000 mm of mercury 2 2mL 356 © John Bird Published by Taylor and Francis EXERCISE 145, Page 319 1. Some gas initially at 16C is heated to 96C at constant pressure. If the initial volume of the gas is 0.8 m 3 , determine the final volume of the gas. Since the change occurs at constant pressure (i.e. an isobaric process), Charles’ law applies, VV i.e. 12 TT12 3 where V 1 = 0.8 m , T 1 = 16C = (16 + 273)K = 289 K and T 2 = (96 + 273)K = 369 K. 0.8 V Hence, 2 289 369 (0.8)(369) from which, volume at 96C, V = = 1.02 m 3 2 289 2. A gas is contained in a vessel of volume 0.02 m 3 at a pressure of 300 kPa and a temperature of 15C. The gas is passed into a vessel of volume 0.015 m 3 . Determine to what temperature the gas must be cooled for the pressure to remain the same. Since the process is isobaric it takes place at constant pressure and hence Charles’ law applies, VV i.e. 12 TT12 3 3 where T 1 = (15 + 273)K = 288 K and V 1 = 0.02 m and V 2 = 0.015 m 0.02 0.015 Hence 288 T2 (0.015)(288) from which, final temperature, T = = 216 K or (216 - 273)C i.e. - 57C 2 0.02 3. In an isobaric process gas at a temperature of 120C has its volume reduced by a sixth. Determine the final temperature of the gas. Since the process is isobaric it takes place at constant pressure and hence Charles’ law applies, 357 © John Bird Published by Taylor and Francis VV i.e. 12 TT12 5 where T = (120 + 273)K = 393 K and V = V 1 2 6 1 5 V V 1 Hence 1 6 393 T2 5 V1 393 6 5 from which, final temperature, T 2 = 393 = 327.5 K V61 = (327.5 - 273)C = 54.5C 4. The volume of a balloon is 30 litres at a temperature of 27ºC. If the balloon is under a constant internal pressure, calculate its volume at a temperature of 12ºC. VV Since 12 TT12 where T1 = (27 + 273)K = 300 K and T2 = (12 + 273)K = 285 K 30litres V Hence, 2 300K 285K 30litres 285K and new volume, V = = 28.5 litres 2 300K 358 © John Bird Published by Taylor and Francis EXERCISE 146, Page 320 1. Gas, initially at a temperature of 27C and pressure 100 kPa, is heated at constant volume until its temperature is 150C. Assuming no loss of gas, determine the final pressure of the gas. pp Since the gas is at constant volume, the pressure law applies, i.e. 12 TT12 where T 1 = (27 + 273)K = 300 K, T 2 = (150 + 273)K = 423 K and p 1 = 100 kPa 100 p Hence, 2 300 423 (100)(423) from which, final pressure, p = = 141 kPa 2 300 2. A pressure vessel is subjected to a gas pressure of 8 atmospheres at a temperature of 15ºC. The vessel can withstand a maximum pressure of 28 atmospheres. Calculate the gas temperature increase the vessel can withstand. PP Since 12 TT12 where T1 = (15 + 273)K = 288 K 8atmospheres 28atmospheres Hence, 288K T2 28atmospheres 288K from which, new temperature, T = = 1008 K or (1008 – 273)ºC = 735ºC 2 8atmospheres Hence, temperature rise = (1008 – 288)K = 720 K or temperature rise = (735 – 15)ºC = 720ºC Note that a temperature change of 720 K = a temperature change of 720ºC 359 © John Bird Published by Taylor and Francis EXERCISE 147, Page 321 1. A gas A in a container exerts a pressure of 120 kPa at a temperature of 20C. Gas B is added to the container and the pressure increases to 300 kPa at the same temperature. Determine the pressure that gas B alone exerts at the same temperature. Initial pressure, p A = 120 kPa, and the pressure of gases A and B together, p = p A + p B = 300 kPa By Dalton's law of partial pressure, the pressure of gas B alone is p B = p - p A = 300 - 120 = 180 kPa 360 © John Bird Published by Taylor and Francis EXERCISE 148, Page 322 1. A gas occupies a volume of 1.20 m 3 when at a pressure of 120 kPa and a temperature of 90C. Determine the volume of the gas at 20C if the pressure is increased to 320 kPa. pV pV Using the combined gas law: 11 2 2 TT12 3 where V 1 = 1.20 m , p 1 = 120 kPa, p 2 = 320 kPa, T 1 = (90 + 273)K = 363 K and (120)(1.20) (320) V T = (20 + 273)K = 293 K, gives: 2 2 363 293 (120)(1.20)(293) from which, volume at 20C, V = = 0.363 m 3 2 (363)(320) 2. A given mass of air occupies a volume of 0.5 m 3 at a pressure of 500 kPa and a temperature of 20C. Find the volume of the air at STP. pV pV Using the combined gas law: 11 2 2 TT12 3 where V 1 = 0.5 m , p 1 = 500 kPa, T 1 = (20 + 273)K = 293 K , p 2 = 101.325 kPa, and (500)(0.5) (101.325) V T = 0C = 273 K, gives: 2 2 293 273 (500)(0.5)(273) from which, volume at STP, V = = 2.30 m 3 2 (293)(101.325) 3. A balloon is under an internal pressure of 110 kPa with a volume of 16 litres at a temperature of 22ºC. If the balloon’s internal pressure decreases to 50 kPa, what will be its volume if the temperature also decreases to 12ºC. 361 © John Bird Published by Taylor and Francis pV pV Using the combined gas law: 11 2 2 TT12 where V 1 = 16 litres, p 1 = 110 kPa, p 2 = 50 kPa, T 1 = (22 + 273)K = 295 K and (110)(16) (50) V T = (12 + 273)K = 285 K, gives: 2 2 295 285 (110)(16)(285) from which, volume at 12C, V = = 34.0 litres 2 (295)(50) 4. A spherical vessel has a diameter of 2.0 m and contains hydrogen at a pressure of 300 kPa and a temperature of - 30C. Determine the mass of hydrogen in the vessel.