10.3 Solutions
453
10.56 For a perfect gas of 1 mol
PV = RT
(1)
Under isothermal conditions T = constant. Differentiating (1)
PdV + V dP = 0
( 2 )
The bulk modulus for the isothermal process
K T = −V
dP
dV
T
= P
(3)
For adiabatic compression in which heat of compression remains in the gas
PV
γ
= constant
(4)
where γ = C p /C v is the ratio of specific heats at constant pressure and
constant volume. Differentiating (4)
γ PV
γ −1 dV + V
γ dP = 0
( 5 )
Thus adiabatic elasticity K H is given by
K H = −V
dP
dV
H
= γ P
(6)
It follows that
K H = γ K T
(7)
The adiabatic elasticity is greater than the isothermal elasticity by a factor γ
which is always greater than unity.
10.57 Y = 2η(1 + σ )
∴
σ =
Y
2η
− 1 =
1
2
× 2.5 − 1 = 0.25
10.58 From Fig. 10.13 the new length L = 2AD = 2
AC
2
+ CD
2
= 2
(0.6) 2 + (0.02) 2 = 1.200666 m
Elongation of the wire, L = L − L = 1.200666 − 1.20 = 0.000666 m.
Strain = L/L = 0.000666/1.2 = 5.55 × 10
−4
453
10.56 For a perfect gas of 1 mol
PV = RT
(1)
Under isothermal conditions T = constant. Differentiating (1)
PdV + V dP = 0
( 2 )
The bulk modulus for the isothermal process
K T = −V
dP
dV
T
= P
(3)
For adiabatic compression in which heat of compression remains in the gas
PV
γ
= constant
(4)
where γ = C p /C v is the ratio of specific heats at constant pressure and
constant volume. Differentiating (4)
γ PV
γ −1 dV + V
γ dP = 0
( 5 )
Thus adiabatic elasticity K H is given by
K H = −V
dP
dV
H
= γ P
(6)
It follows that
K H = γ K T
(7)
The adiabatic elasticity is greater than the isothermal elasticity by a factor γ
which is always greater than unity.
10.57 Y = 2η(1 + σ )
∴
σ =
Y
2η
− 1 =
1
2
× 2.5 − 1 = 0.25
10.58 From Fig. 10.13 the new length L = 2AD = 2
AC
2
+ CD
2
= 2
(0.6) 2 + (0.02) 2 = 1.200666 m
Elongation of the wire, L = L − L = 1.200666 − 1.20 = 0.000666 m.
Strain = L/L = 0.000666/1.2 = 5.55 × 10
−4