ФЫВАПРОЫВАПРОЛЯЧСМИТАЧСМИТЬБДЗ А П Р ОЛДЖЫВАПРСМИТЬБЮ ЯЧСМИ 1DON'T USE AI | DON'T USE AI |DON'T USE AI | DON'T USE AI |DON'T USE AISOLVE STEP BY STEP IN DIGITALDetermine the specific heat of the followingsubstances| SubstanceI Acetylene| AcetyleneIn-Butane| AirTemperature (K) | cp (kJ/kmol. K) |-III|-| 300.15 к| 517.15 к| 300.15 К| 300.15 кFORMAT

Answered: Determine the specific heat of the…

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

Chapter1: Introduction

Section: Chapter Questions

Problem 1.1P

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ФЫВАПРОЫВАПРОЛ
ЯЧСМИТАЧСМИТЬБ
ДЗ А П Р ОЛДЖЫВАПР
СМИТЬБЮ ЯЧСМИ 1
DON'T USE AI | DON'T USE AI |DON'T USE AI | DON'T USE AI |DON'T USE AI
SOLVE STEP BY STEP IN DIGITAL
Determine the specific heat of the following
substances
| Substance
I Acetylene
| Acetylene
In-Butane
| Air
Temperature (K) | cp (kJ/kmol. K) |
-I
I
I
|-
| 300.15 к
| 517.15 к
| 300.15 К
| 300.15 к
FORMAT

Expert Solution

Step 1: Specific heat of Acetylene

Specific heat of Acetylene is given by

$C subscript P over R equals A space plus space B T space plus space C T squared plus D over T squared R space i s space u n i v e r s a l space g a s space c o n s tan t C subscript P over R equals 6.132 space plus 1.952 asterisk times 10 to the power of negative 3 end exponent T minus 129900 over T squared T h u s comma space S p e c i f i c space h e a t space a t space 300.15 space K space i s fraction numerator C subscript P over denominator 8.314 end fraction equals 6.132 plus 1.952 asterisk times 0.001 asterisk times 300.15 space minus fraction numerator 129900 over denominator 300.15 squared end fraction space C subscript P equals 8.314 asterisk times 5.276 equals 43.86 space fraction numerator k J over denominator k m o l space K end fraction S i m i l a r l y comma space a t space 517.15 space K space t e m p e r a t u r e fraction numerator C subscript P over denominator 8.314 end fraction equals 6.132 plus 1.952 asterisk times 0.001 asterisk times 517.15 space minus fraction numerator 129900 over denominator 517.15 squared end fraction C subscript P equals 8.314 asterisk times 6.656 equals 55.34 space fraction numerator k J over denominator k m o l space K end fraction$