Question

2.50 mole of perfect gas at 0°C is expanded from an initial pressure of 5.00 atm to a final pressure of 1.00 atm against a constant external pressure of 1.00 atm. (a) isothermally and (b) adiabatically. (1) Calculate the values of q, w, DU, DH, DS, DSsur and DStot for (a) and (b) (2) Draw p -V diagram to represent these two different paths.

Answer 1

ISothermal expansion;
w=-2.303nRTlog[p1/p2]
T=0+273k=273k
p1=5atm,p2=1atm
number of moles n=2.5moles
R=0.0821Latm/K/mol

w=-2.303*2.5*0.0821*273*log5/1
-90.198Latm
heat q= -[work done]
=-[-90.198]=90.198Latm

in isothermal expansion there is no change in temperature ie
DT=0
change in internal energy is also zero for isothermal expansion.ie
DU=0
DH is the change in enthalpy
DH=DU-w
DH=-w=90.198
DS =Q/T=90.198/273=0.33Latm/K
DSsur=-[DS]=-0.33latm/k
DStotal=DS+DSsur=0

B,Adiabatic expansion;

Q=0 because there is no heat is absorbed or released
DU=workdone w
w=-2.303nRTLOG[P1/P2]=-90.198lATM
SO DU=-90.198lATM
DH=0
DS=Q/T=0,IT IS ALWAYS ZER0 FOR ADIABATIC PROCESS
THEREFORE DSsur=0
DStotal=0

you will get a steeper curve in P-V diagram of adiabatic process