Molar Mass and Avogadro’s Number - General Physics II | PHY 212, Exams of Physics

Download Molar Mass and Avogadro’s Number - General Physics II | PHY 212 and more Exams Physics in PDF only on Docsity! Phy 212: General Physics II 1 Chapter 19 Worksheet Boltzmann Constant: k = 1.38x10-23 J/K Gas Constant: R = 8.314 J/mol.K Molar Mass & Avogadro’s Number: 1. Determine the atomic mass for the following elements: {from the textbook} a) hydrogen (H): 1.00797 g/mol b) argon (Ar): 39.948 g/mol c) carbon (C): 12.01115 g/mol d) oxygen (O): 15.9994 g/mol 2. Determine the molecular mass for the following compounds: a) oxygen gas (O2): 31.9988 g/mol b) carbon dioxide (CO2): 44.00995 g/mol c) methane (CH4): 16.4303 g/mol 3. Determine the molar quantity of the following: a) 0.10 kg of Ar: 2.5 mol b) 0.010 kg of O2: 3.1 mol c) 0.15 kg of CO2: 3.4 mol d) 1.0 kg methane (CH4): 62.3 mol 4. Determine the number of molecules for the following: a) 0.10 kg of Ar: 1.5x1024 atoms b) 0.010 kg of O2: 1.9x10 24 molecules c) 0.15 kg of CO2: 2.1x10 24 molecules d) 1.0 kg methane (CH4): 3.8x10 25 molecules 5. A sample of monatomic ideal gas is originally at 20°C. What is the final temperature of the gas if both the pressure and volume are doubled? Ans. ( )2 2 1 1 2 2 1 12 1 2 1 1 1 1 1 P V P V P V 4P V = T = T = 293.15K = 1172.6K T T P V P V     ⇒         6. Heat is absorbed by a sample of a monatomic ideal gas at 40 °C. It is observed that the gas expands until its volume is doubled and the pressure drops to half of its original value. What is the final temperature of the gas? Ans. ( ) 1 1 2 2 1 1 2 2 2 1 2 1 1 1 1 1 P 2V P V P V P V 2 = T = T = 313.15K = 313.15K T T P V P V   ⋅   ⇒          Phy 212: General Physics II 2 Chapter 19 Worksheet Boltzmann Constant: k = 1.38x10-23 J/K Gas Constant: R = 8.314 J/mol.K 7. A canister containing 150 kg of an ideal gas has a volume of 8.0 m3. If the gas exerts a pressure of 5.0 × 105 Pa, what is the rms speed of the molecules? Ans. ( )( )5 3 6 mrms s 3nRT 3PV PV = nRT = 5.0 10 Pa 8.0m =4.0 10 J v = = = 282.8 m m × × ⇒ 8. A tank contains 135 moles of the monatomic gas, argon, at a temperature of 15.3 °C. How much energy must be added to the gas to increase its temperature to 45.0 °C? Ans. ( )( )( ) 43 3 J2 2 molKQ = nR T = 135mol 8.314 29.7K =5.00 10 J⋅∆ × 9. An ideal gas with a fixed number of molecules is maintained at a constant pressure. At 30.0 °C, the volume of the gas is 1.50 m3. What is the volume of the gas when the temperature is increased to 75.0 °C? Ans. ( ) 3 32 1 1 2 2 2 1 1 V V V 1.5m = V = T = 348.15K = 1.7m T T T 303.15K     ⇒        Constant Temperature (Isothermal) Processes: 1. A sample of 10.0 moles of a monatomic ideal gas, held at constant temperature (1000K), is expanded from 0.10 m3 to 0.20 m3. a. What is the initial pressure of the gas, before heating? Ans. ( )( )( ) ( ) J molK 5 3 10mol 8.314 1000KnRT P = = = 8.31 10 Pa V 0.10m ⋅ × b. What is the initial internal energy of the gas? Ans. ( )( )( ) 53 3 Jint 2 2 molKE = nRT = 10mol 8.314 1000K =1.25 10 J⋅ × c. How much work is performed by the gas during the heating process? Ans. ( )( )( ) ( ) 42 JmolK 1 V W = nRT ln = 10.0mol 8.314 1000K ln 2 = 5.76 10 J V ⋅   ⋅ ⋅ ×    d. What is the final internal energy of the gas? Ans. Since ∆T=0 K, i f 5 int intE = E = 1.25 10 J× e. What is the change in internal energy of the gas? Ans. Eint=0 J f. Conceptually, is heat absorbed or released by the gas during the compression? Ans. Q is absorbed since the gas does work but Eint doesn’t change. g. Calculate the amount of heat absorbed/released by the gas? Ans. intE = Q - W = 0 J∆ or 4Q = W = 5.76 10 J× Phy 212: General Physics II 5 Chapter 19 Worksheet Boltzmann Constant: k = 1.38x10-23 J/K Gas Constant: R = 8.314 J/mol.K c. How much work is performed on the gas during the heating process? Ans. W = 0 J d. What is the final internal energy of the gas? Ans. ( )( )( ) 45 5 Jint 2 2 molKE = nRT = 10.0 mol 8.314 400. K = 8.31 10 J⋅ × e. What is the change in internal energy of the gas? Ans. ( )( )( ) 45 5 Jint 2 2 molKE = nR T = 10.0 mol 8.314 150. K = 3.12 10 J⋅∆ ∆ × f. Conceptually, is heat absorbed or released by the gas during the compression? Ans. Since Eint increases, Q must be absorbed to account for the additional energy. g. Calculate the amount of heat absorbed/released by the gas? Ans. 4 intQ = E = 3.12 10 J∆ × Adiabatic Processes: 5. A sample of 1.5 moles of a thermally insulated (adiabatic) monatomic ideal gas is expanded from 0.10 m3 to 0.20 m3. During the expansion, the pressure of the gas decreases from 3.039x105 Pa to 0.957x105 Pa a. What is the initial temperature of the gas, before adiabatic expansion? Ans. ( )( ) ( )( ) 3 5 i J molK 0.10m 3.39 10 PaPV T = = = 2440K nR 1.5mol 8.314 ⋅ × b. What is the final temperature of the gas, after adiabatic expansion? Ans. ( )( ) ( )( ) 5 3 f J molK 0.957 10 Pa 0.20mPV T = = = 1530K nR 1.5mol 8.314 ⋅ × c. What is the change in internal energy of the gas? Ans. ( )( )( ) 43 3 Jint 2 2 molKE = nR T = 1.5 mol 8.314 -910. K = -1.70 10 J⋅∆ ∆ × d. How much heat is gained by the gas during this process? Ans. Q = 0 J (adiabatic process!) e. How much work is performed by the gas during the heating process? Ans. 4 intW = - E = 1.70 10 J∆ × Phy 212: General Physics II 6 Chapter 19 Worksheet Boltzmann Constant: k = 1.38x10-23 J/K Gas Constant: R = 8.314 J/mol.K 6. A sample of 5.0 moles of a thermally insulated (adiabatic) monatomic ideal gas is expanded from 0.10 m3 to 0.20 m3. During the expansion, the temperature of the gas decreases from 800 K to 500 K. a. What is the initial pressure of the gas, before adiabatic expansion? Ans. ( )( )( ) ( ) J molK 5 i 3 5.0mol 8.314 800KnRT P = = = 3.33 10 Pa V 0.10m ⋅ × Phy 212: General Physics II 7 Chapter 19 Worksheet Boltzmann Constant: k = 1.38x10-23 J/K Gas Constant: R = 8.314 J/mol.K b. What is the final pressure of the gas, after adiabatic expansion? Ans. ( )( )( ) ( ) J molK 5 f 3 5.0mol 8.314 500KnRT P = = = 1.04 10 Pa V 0.20m ⋅ × c. What is the change in internal energy of the gas? Ans. ( )( )( ) 43 3 Jint 2 2 molKE = nR T = 5.0 mol 8.314 -300. K = -1.87 10 J⋅∆ ∆ × d. How much work is performed by the gas during the process? Ans. 4 intW =- E = 1.87 10 J∆ ×