Question

25 pts) One mole of ideal, monatomic gas, initially at T = 250 K and pressure 5.0 atm is: a) reversibly heated at constant pressure until its volume doubles b) reversibly heated at constant volume until its pressure doubles Determine w, q ,ΔU, ΔΗ , and ΔS for these two cases (20 pts). Can you calculate A and G for these two cases? Explain why. (5 pts)

Answer 1

Given:
Ti = 250 K
n= 1 mol
Pi = 5 atm
Vi = n*R*T / P
    = 1*0.0821*250 / 5
    =4.105 L
For monoatomic gas:
CV = 3*R/2 = (3*8.314)/2 = 12.5 JK-1mol-1
CP = 5*R/2 = (5*8.314)/2 = 20.8 JK-1mol-1

a)
Vf =2*Vi = 2*4.105 L = 8.21 L
When pressure is constant, it is isobaric process
Vi/Ti = Vf/Tf
so, when V is doubled, T will also double
Tf= 2*Ti = 2*250 K = 500 K

W= -P (Vf-Vi)
    = -5 atm* (8.21 - 4.105) L
    = -20.525 atm-L
     = -2079.2 J

Q = n*Cp*delta T
   = 1*20.8*(Tf-Ti)
   = 20.8*(500 - 250)
= 5200 J

delta H =q = 5200J

delta U = n*Cv*(Tf-Ti)
              = 1*12.5*(500-250)
              = 3125 J

delta S = n*CP*ln(Tf/Ti)
              = 1*20.8*ln (500/250)
              = 14.42 Jmol-1K-1

b)
Pf =2*Pi = 2*5 atm = 10 atm
When Volume is constant, it is isochoric process
Pi/Ti = Pf/Tf
so, when P is doubled, T will also double
Tf= 2*Ti = 2*250 K = 500 K

W= -P (Vf-Vi)
    = 0

Q = n*Cp*delta T
   = 1*20.8*(Tf-Ti)
   = 20.8*(500 - 250)
= 5200 J

delta H =q = 5200J

delta U = Q+ W
               = Q
                = 5200 J

delta S = n*Cv*ln(Tf/Ti)
              = 1*12.5*ln (500/250)
              = 8.66 Jmol-1K-1

A and G can be calculated at constant temperature only