3. For an ideal gas under adiabatic reversible processes, i ts volume and tem pe

Question

3. For an ideal gas under adiabatic reversible processes, i ts volume and tem perature follow the relation: constant. .m/R and Cv.m is the molar heat capacity at constant volume and R is the gas (a) (8pt) The vol compression t constant volume Cv.m ume of a 1.00-mol ideal gas initially at 106.06 K is 49.236 L. After the adiabatic reversible 24.618 L, the temperature becomes 300.0 K. Determine the molar heat capacity at pt) complete the following thermodynamic quantities (in Joule or in KJ) for the above process. Show your detailed calculations AS AU (c) (5pt) What is the possible unit for entropy?

Explanation / Answer

For adiabatic reversible compression,

V1T1C= V2T2C,

(V1/V2)= (T2/T1)C, C= CV/R

Where T2= temperature after compression =300K, V2= volume after compression = 24.618L

T1= temperature before compression =106.06K, V1= volume before compression =49.236L, R = 8.314 J/mole.K

49.236/24.618= (300/100.06)C,

2= 3C

Taking ln, ln 2= C*ln3

C= ln2/ln3= 0.63

CV/R= 0.63, CV= 8.314*0.63= 5.24 J/mole.K

CP= CV+R= 5.24+8.314= 13.554

CP/CV= Y= 13.554/5.24= 2.59

For adiabatic compression, Q=0

From 1st law of thermodynamics, deltaU= Q+W becomes

deltaU= change in internal energy= W

W= n*R*(T2-T1)/ (Y-1)

T2=300 and T1= 106.06, n= no of moles= 1

W= 1* 8.314*(300-106.06)/ (2.59-1) =1014Joules

For adiabatic reversible compression, deltaS= entropy change =0

deltaH= n*Cp*(T2-T1)= 1*13.554*(300-106.06)=2629 joules

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