2.20 mol of N3 at 12.0 °C and 6.45 bar undergoes a transformation to the state described by 335 "Cand 1.95 bar.Part ACalculate ASCr3/ K)- 30.81 - 11.87 x 10 +2.3968 x 10--1.0176 x 10-Express your answer with the appropriate units.JAS - 48.86K

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2.20 mol of N, at 12.0"C and 6.45 bar undergoes a transformation to the state described by 335 "C and 1.95 bar. Part A Calculate AS -30.81 - 11.87 x 1o+2.3968 × 10-*- 1.0176 x 10E 1x 104 Express your answer with the appropriate units

2.20 mol of N, at 12.0"C and 6.45 bar undergoes a transformation to the state described by 335 "C and 1.95 bar. Part A Calculate AS -30.81 - 11.87 x 1o+2.3968 × 10-*- 1.0176 x 10E 1x 104 Express your answer with the appropriate units

Physical Chemistry

2nd Edition

ISBN:9781133958437

Author:Ball, David W. (david Warren), BAER, Tomas

Publisher:Ball, David W. (david Warren), BAER, Tomas

Chapter1: Gases And The Zeroth Law Of Thermodynamics

Section: Chapter Questions

Problem 1.66E

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Question

2.20 mol of N3 at 12.0 °C and 6.45 bar undergoes a transformation to the state described by 335 "Cand 1.95 bar.
Part A
Calculate AS
Cr
3/ K)
- 30.81 - 11.87 x 10 +2.3968 x 10--1.0176 x 10-
Express your answer with the appropriate units.
J
AS - 48.86
K

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