1. No category

A sample consisting of 2 mol of He is expanded isothermally at
o
22 C from 22.8 L to 31.7 L calculate the work reversibly.
n =2
T = 22oC = 295 K
V1 = 22.8 L
V2 = 31.7 L
W =???
Work at isothermal reversible expansion:
W = -nRTLn(V2/V1)
= - 2× 1.987 × 295 Ln(31.7/22.8) = -386.3 cal
When one mole of liquid Br2 is converted to Br2 vapour at 25oC and 1 atm
pressure, 7.3 Kcal of heat is absorbed and 0.59 kcal of expansion work is
done by the system. Calculate ΔE for this process.
n =1
T = 25oC = 298 K
q = 7.3 kcal
P = 1 atm
w = - 0.59 kcal
ΔE = q + w = +7.3 – 0.59 = 6.71 kcal
ΔE =???
Calculate w for the adiabatic reversible expansion of 3 moles of an ideal
gas at 275 K and 20 atm to a final pressure of 2 atm. (γ = 5/3)
n =3 mol
T = 275 K
P1 = 20 atm
P2 = 2 atm
γ = 5/3
W =???
Work at adiabatic reversible expansion:
w
 nR(T2  T1 )
1 

 T2 
P 
    2 
 T1 
 P1 
T2 = ???
 1
5
3
5
1
3
5
3
2
3
 T2 
 2 

  
 275 
 20 
 T2 
 2 

  
 275 
 20 
5  T2  2
ln 
  ln 0.1  1.535
3  275  3
T2  109 .47 K
 nR(T2  T1 )
1 
3  8.314  (109 .47  275 )
3  8.314  (109 .47  275 )
w

5
2
1

3
3
3  3  8.314  (109 .47  275 )
w
 6.16 kJ
2
w
When 2 mol of CO2 is heated at a constant pressure of 1.25 atm.
Its temperature increase from 260 K to 277 K. Given that the
molar heat capacities of CO2 at constant pressure is 37.11
J/K.mol, calculate q, ΔH and ΔE.
n =2 mol
T1 = 260 K
T2 = 277 K
q p  H  nCP (T2  T1 )
H  2  37 .11(277  260 )  1261 .7 J
E  nCV (T2  T1 )
E  n(CP  R)(T2  T1 )
H  2  (37 .11  8.314 )(277  260 )  979 .06 J
look at the solved problems in the reference.
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