a) for reversile adiabatic expansion,q=0 , W= nR*(T2-T1)/ (Y-1), Y= CP/CV= ratio of specific heats of Ne=1.67
n= no of moles of gas
R= 8.314 Joules/mole.K, Temperature( T2) at the end of expansion process is
T2/T1= (V1/V2) R/CV= (V1/V2) (Y-1)
T1= 75+273= 348 K, V1= V and V2=2V
T2= 348*(1/2)(1.67-1)=218.7 K
hecne work done = 1*8.314*(218.7-348)/0.67=-1604 Joules, work is -ve since system does work on the surroundings
since from 1st law of thermodynamics, deltaU = q+W
deltaU= change in internal energy= -1604 joules for neonm CP= 20.78 J/mole.K
deltaH= n*CP*temperature difference= 1*20.78* (218.7-348)= -2687 Joules
since the expansion is adiabatic, P1V1Y= P2V2Y
2*VY= P2*(2V)Y
P2 = 2*(1/2)Y = 0.63 atm
b) for isothermal compression of an ideal gas, deltaU=0, deltaH=0 and deltaU=W= nRTln(V1/V2)
=1*8.314*348*ln(1/2)= -2005 joules
c) for constant pressure ( isobaric expansion)
from gas law P1V1/T1= P2V2/T2
gvien V2= 2V1
P1= 2atm, T1= 348K, P2= 2atm, V1=V and V2=2V
T2= (V2/V1)*T1=2*348= 696K
work done = -nR*(T2-T1)=-8.314*(696-348)= -2893 joules
deltaH= nCP*dT=1*20.78*(696-348)= 7231 Joules
deltaU= nCv*(T2-T1)= (20.78-8.314)*(696-348)=4338 joules
q=deltaH=7231 joules
4. work done= -Pext.dV, dV= volume change, volume initially, V1= nRT/P= 1*0.0821 L.atm/mole.K* 348/2= 14.28 L
volume after expansion =2*14.28= 28.56L
work done= -0.5*(28.56-14.28)L.atm. 101.3J/L.atm =-723 joules
for adiabatic, q=0 and deltaU= -723 joules
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