Under constant pressure P = 1 atm an ideal gas of 1.5 mol undergoes cooling from 120°C to 30.0°C, = 3/2R. Please calculate q, w, AU, and AH (all units are J). and Cv,m Please enter all four answers using scientific notation (E notation) with three significant figures, such as 2.30E4 (=23000). If the answer is negative, please do not forget the negative sign. If answer is zero, please just enter 0 without decimal. (1) q= (2) w= (3) AU= (4) ΔΗ=

Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
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QUESTION 2
Under constant pressure P = 1 atm an ideal gas of 1.5 mol undergoes cooling from 120°C to 30.0°C,
and = 3/2R. Please calculate q, w, AU, and AH (all units are J).
Cv.m
Please enter all four answers using scientific notation (E notation) with three significant figures, such as
2.30E4 (=23000). If the answer is negative, please do not forget the negative sign. If answer is zero,
please just enter 0 without decimal.
(1) q=
(2) w=
(3) AU=
(4) ΔΗ=
Transcribed Image Text:QUESTION 2 Under constant pressure P = 1 atm an ideal gas of 1.5 mol undergoes cooling from 120°C to 30.0°C, and = 3/2R. Please calculate q, w, AU, and AH (all units are J). Cv.m Please enter all four answers using scientific notation (E notation) with three significant figures, such as 2.30E4 (=23000). If the answer is negative, please do not forget the negative sign. If answer is zero, please just enter 0 without decimal. (1) q= (2) w= (3) AU= (4) ΔΗ=
Expert Solution
Step 1

Given Details:

At constant Pressure , Gas is being cooled

P=1 atm

no. of moles of gas ,n=1.5 mol

Initial Temperature, T1=1200C=120+273=393 K

Final Temperature, T2=30.00C=30+273=303 K

Cv,m=3R2

we need to determine the q,W,U and H.

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