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Part A Starting with 2.80 mol of No gas (assumed to be ideal) in a cylinder at 1.00 atm and 19.0"C, a chemist first heats the gas at constant volume, adding 1.52 x 10* J of heat, then Calculate the final temperature of the gas. continues heating and allows the gas to expand at constant Express your answer in kelvins. pressure to twice its original volume ? Trinal = K Submit Request Answer Part B Calculate the amount of work done by the gas Express your answer with the appropriate units. W : Value Units Submit Request Answer Part C Calculate the amount of heat added to the gas while it was expanding. Express your answer with the appropriate units. 0 0 ? Q = Value Units Submit Request Answer Part D Calculate the change in internal energy of the gas for the whole process. Express your answer with the appropriate units. I'MA AU = Value Units P PearsonPart A Calculate the final temperature of the gas. Express your answer in kelvins. AEd Trinal = 1168.62 K Submit Previous Answers Request Answer X Incorrect; Try Again; 9 attempts remaininggiven information, in the coustion . initial pressure = P, = 1.0 atm = 1. 01325 x 105 Pa . initial temperature = T. = 196 = 19+ 273 = 292k . Heat ( Q ) = 1052 x 10 YJ . cevanity ( n ) = 2. 80 mol for a gas ideal "like Nitrogen the volumentric + Pressure specific heat is equal to CV = [ X R = CV: 5 x 8. 314 = Cv = 0. 8 J/ mol.K Cp = F X R = > V, = MRT, P , v, = ( 2 . 80 mol ) X ( 8 . 314 J / K . mol ) x ( 292 K ) ) (9 1. 013 25 X 10 Pa V = 0 . 0671. m3 0 = ncr ( T 2 - T. ) => 72 = + T, HOW T2= 1.52 x101 + 292 2 . 80 X 20 . 8 T2 = 260 . 99 + 292 T2- 553 K for constant volume heating, I- constantP 2 ( 1 atm ) 1 2 T 1 292 k 553 K 12 = 1894 atm for constant pressure heatingV- constant V 2 V2 - 2 V, given in Question T 3 0. 0671 2x 0.0671 To = 2x 553 2 5 53 73= 1106 K B) for constant volume heating work tone. = pov = 0 ( constant volume ) tomos . ( ) = V for constant pressure heating, work done = POV =P2 * ( 12-4) work fone = / 1 . 894 x 10 01325 x10 / x ( 20 0671 - 0.0671 ) work Jone = 12, 877 . 13 J () For constant pressure Process, change in internal energy = mcv (T 3 - T2 ) = 2 . 80 20 . 8 x (1106 - 553 ) = 32, 206 - 72 J Q - W = change in infernal energy lead role c = w+ change in infernal energyQ = 1 2, B77 . 13 J + 32, 206 . 72 ] Q = 45,0 B3 . 85 J D for the whole process, change in infernal energy AU = mer (T3 - Tz) = 2 80 X 20- 8 x ( 1106 - 292 ) AU = 47, 407 . 36 J )