Physics 41 Chapter 21 Lecture Problems −23 23 PV== nRT NkT, Q =∆= mc T, R8.31 J / mol ⋅= K , k1.38 x 10 J / K , NA =6.022 x 10 / mol 5 −27 382kTBBB kT kT 1atm= 1.013 x 10 Pa , u = 1.66 x 10 kg, vrms = ,vavg = ,vmp = mooπ mm 1. A 5.00-L vessel contains nitrogen gas at 27.0°C and 3.00 atm. Find (a) the total translational kinetic energy of the gas molecules and (b) the average kinetic energy per molecule. Nmv2 Nmv2 (a) PV= nRT = The total translational kinetic energy is = E : 3 2 trans 33 53− Etrans = PV =(3.00 ×× 1.013 10)( 5.00 ×= 10) 2.28 kJ 22 mv2 3kT 3RT 3( 8.314)( 300) (b) =B = = =6.21 × 10−21 J 23 2 22N A 2( 6.02× 10 ) 2. Calculate the RMS speed of an oxygen molecule in the air if the temperature is 5.00 °C. The mass of an oxygen molecule is 32.00 u −23 3kT 3(1.38x 10 JK / )278 K = vrms = = 466ms / m (32u )(1.66 x 10−27 kg / u ) 3. A cylinder contains a mixture of helium and argon gas in equilibrium at 150°C. (a) What is the average kinetic energy for each type of gas molecule? (b) What is the root-mean- square speed of each type of molecule? 33 −−23 21 (a) K= kTB =(1.38×= 10 J K)( 423 K) 8.76× 10 J 22 1 (b) K= mv2 =8.76 × 10−21 J 2 rms 1.75× 10−20 J so v = (1) rms m 4.00 g mol For helium, m = =6.64 × 10−24 g mol ecul e 6.02× 1023 mol ecul es mol m =6.64 × 10−27 kg mol ecul e 39.9 g mol Similarly for argon, m = =6.63 × 10−23 g mol ecul e 6.02× 1023 mol ecul es mol m =6.63 × 10−26 kg mol ecul e Substituting in (1) above, we find for helium, vrms = 1.62 km s and for argon, vrms = 514 m s 4. In a constant-volume process, 209 J of energy is transferred by heat to 1.00 mol of an ideal monatomic gas initially at 300 K. Find (a) the increase in internal energy of the gas, (b) the work done on it, and (c) its final temperature n = 1.00 mol , Ti = 300 K (b) Since V = constant, W = 0 (a) ∆Eint = QW + =209 J += 0 209 J 3 (c) ∆E = nC ∆= T n R ∆ T int V 2 2∆E 2( 209 J) so ∆=T int = =16.8 K 3nR 3( 1.00 mol) ( 8.314 J mol⋅ K ) TT=i +∆ T =300 K + 16.8 K = 317 K 5. A 2.00-mol sample of a diatomic ideal gas expands slowly and adiabatically from a pressure of 5.00 atm and a volume of 12.0 L to a final volume of 30.0 L. (a) What is the final pressure of the gas? (b) What are the initial and final temperatures? (c) Find Q, W, and ∆Eint. γγ (a) PVifi = P Vf γ V 12.0 1.40 PP=i = 5.00 atm = 1.39 atm fi V f 30.0 5 −33 PV 5.00( 1.013×× 10 Pa)( 12.0 10 m ) (b) T =ii = = 365 K i nR 2.00 mol( 8.314 J mol⋅ K ) 5 −33 PVff 1.39( 1.013×× 10 Pa)( 30.0 10 m ) T = = = 253 K f nR 2.00 mol( 8.314 J mol⋅ K ) (c) The process is adiabatic: Q = 0 CP RC+ V 5 γ =1.40 = = , CRV = CCVV 2 5 ∆E = nC ∆= T 2.00 mol 8.314 J mol⋅ K( 253 K − 365 K) =− 4.66 kJ int V ( ) 2 WEQ=∆int − =−4.66 kJ −=− 0 4.66 kJ .