Lecture 4, First Law Calc 1-14-08

# Lecture 4 First - Example – Gas heated at constant volume 10.0 liters of helium gas at STP heated to 375 K at constant V Calculate ∆ E Closed

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Unformatted text preview: Example – Gas heated at constant volume 10.0 liters of helium gas at STP heated to 375 K at constant V. Calculate ∆ E Closed System 10.0 liters He 273 K 1.00 atm q is + Surroundings 375 K heat, q Initial State Closed System 10.0 liters He 375 K P increases Δ E is + Surroundings 375 K Final State mol K J 47 . 12 R 2 3 C v ⋅ = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = V V q E w w q E = = + = Δ Δ Example – Gas heated at constant volume Calculation of ∆ E NOTE: CHANGES ARE ALWAYS CALCULATED AS FINAL – INITIAL ∆T = T(FINAL) – T(INITIAL) ∆E = E(FINAL) – E(INITIAL) ( ) ( ) ( ) J q E J q K K mol K J mol q T nC q mol l mol l He of mol V V V v V 568 568 273 375 47 . 12 446 . 446 . 4 . 22 00 . 1 . 10 = = Δ = − ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ = Δ = = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = Write this down for future reference to constant pressure process Heat capacities of gases at constant volume 32.0 Br 2 12.5 Xe 24.1 Cl 2 12.5 Kr 19.9 N 2 12.5 Ar 20.2 CO 12.5 Ne 20.18 H 2 12.5 He C v ( J K-1 mol-1 ), Diatomic gas C v ( J K-1 mol-1 ), Monatomic gas Internal Energy and Temperature • Heat capacities, q and Δ E 1 mole of He, C v = 12.5 J/K·mol T initial = 298 K 1 mole H 2 , C v = 20.18 J/K·mol T initial = 298 K q=25 J q=40.36 J He, T final = 300 K Δ E = 25 J H 2 , T final = 300 K Δ E = 40.36 J Constant V Constant V Δ T is the same for both processes, but heating He by 2 K requires less energy than heating H 2 by 2 K....
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## This note was uploaded on 08/06/2008 for the course CHEM 1B taught by Professor Watts during the Winter '08 term at UCSB.

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Lecture 4 First - Example – Gas heated at constant volume 10.0 liters of helium gas at STP heated to 375 K at constant V Calculate ∆ E Closed

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