A monatomic, ideal gas is in a sealed container (the number of gas molecules is
ID: 2215286 • Letter: A
Question
A monatomic, ideal gas is in a sealed container (the number of gas molecules is always constant: n = 2.0 moles); the initial pressure is Pi = 101000.0 Pa and the initial volume is Vi = 0.0224 m3. First, the volume of the gas is decreased at a constant pressure (at Pi = 101000.0 Pa) to a final volume of Vf = 0.0155 m3. Second, the pressure of the gas is increased at a constant volume (at Vf = 0.0155 m3) to a final pressure of Pf = 135000.0 Pa. How much heat was added to (give as a positive number) or removed from (give as a negative number) the system? (The gas constant R = 8.31 J/mole-K.) Q =Explanation / Answer
Pi = 1.01 10^5 Pa Vi = 0.0224 m^3 Pi Vi = n R Ti Ti = Pi Vi / (n R) Ti = 136.12 K First: Pi = 1.01 10^5 Pa = constant Vf = 0.0155 m^3 Charles's law P = constant Vi / Ti = Vf / Tf Tf = Ti Vf / Vi Tf = 94.19 K Second: Vf = 0.0155 m^3 = constant Pf = 1.35 10^5 Pa Gay-Lussac's law Gay-Lussac's law V = constant Pi/Tf = Pf/T2 T2 = Tf Pf / Pi T2 = 125.89 K First case: An isobaric process occurs at constant pressure http://en.wikipedia.org/wiki/Isobaric_process Qisobaric = n Cp ?T = n Cp (Tf - Ti) Cp = 2.5 R for monoatomic gaz Qisobaric = 2 * 2.5 * 8.31 (94.19 - 136.12) Qisobaric = -1742.2 J the system loses energy Second case: An isochoric process, or isometric/isovolumetric process, occurs at constant volume. http://en.wikipedia.org/wiki/Isochoric_process Qisochoric = n Cv ?T = n Cv (T2 - Tf) 125.89 Cv =1.5 R for monoatomic gases Qisochoric = n Cv ?T = 2 1.5 R ( 125.89 - 94.19 ) Qisochoric= 790.28 J the system received energy