4.1) A perfect gas has a constant volume molar heat capacity of CV ,m 1.5 R and

Question

4.1) A perfect gas has a constant volume molar heat capacity of CV ,m 1.5 R and a constant pressuremolarheatcapacityofCp,m 2.5R.Fortheprocessofheating2.80molofthisgaswitha 120 W heater for 65 seconds, calculate a) q, w, T, and U for heating at a constant volume, b) q, w, T, and H for heating at a constant pressure.

4.2) Determine the heat capacity Cp and the molar heat capacity Cp,m of a solid sample from the observation that transferring the sample with n = 3.65 mol from boiling water (at constant ambient pressure) into 165.0 g of water initially at 25.2oC raised the water temperature to 63.8oC. Assume that no heat is lost to the environment and that water has a constant molar heat capacity of Cp,m = 75.3 J K1 mol1 in the 20oC to 100oC range.

Explanation / Answer

a) at constant volume (isobaric process)

Work done =0

change in internal energy= 120Joules/sec*65 sec=7800 joules

for ischoric ( constant volume process),

from first law of thermodynamics, delU= Q+W

W=0 and hence delU= Q= 7800 joules

delU= moles of gas*Cv*delT

where Cv=1.5R= 1.5*8.314=12.471 joules/mole.K

7800= 2.8*12.471*delT

delT= 7800/(2.8*12.471)=223.3754

b)

For constant pressure process,Q = delH= moles *Cp*delT=2.5*2.5*8.314*223.3754=11601.95 Joules

delU= moles*Cv*delT= 7800 joules

W= -moles* R*delT= -2.5*8.314*223.3754=-4640.7 Joules

4.2

Heat taken up by water= moles of water* specific heat of water (J/k.mol)* delT

Moles of water= mass/molecular weight= 165/18=9.2

Heat taken by water= 9.2*75.3*(63.8-25.2)=26740.54 joules

The solid sample at 100 deg.c initially ( from boiling water) reaches an equilibrium temperature of 63.8 deg.c

heat lost by sample= 3.65*(100-63.8)* Cp= 26740.54

Cp =26740.54/(3.65*36.2)=202.3805 j/mol.K

heat capacity= 26740.54/3.65=7326.174 J/K.

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