Question

Consider the expansion of 1.00 mole of (ideal) Ne from 2.00 atm at 75.00 C; the volume is doubled in the process. Find q, w, DH and the final pressure and temperature for a:

(a) reversible adiabatic expansion

(b) reversible isothermal expansion

(c) reversible constant pressure expansion

(d) irreversible adiabat against 0.500 atm external pressure

Answer 1

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*V^Y= 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