10.3 Solutions
441
∴
15
100
mgh = mcT × 4.18 = m × 1000T × 4.18
∴ T =
15 × 9.8 × 25
100 × 1000 × 4.18
= 0.0088
◦ C
10.40 Mechanical energy available, W = mgh
Heat absorbed, H = mcT J.
Loss of mechanical energy = gain of heat energy
∴
mgh = mcT J
∴
T =
gh
c J
=
9.8 × 100
30.6 × 4.18
= 7.66
◦ C = 7.66 K
10.3.5 Thermodynamics
10.41 (a) (i) In an isobaric process pressure remains constant.
(ii) In an isochoric process volume remains constant.
(iii) In an adiabatic process heat is neither absorbed nor evolved by the
system.
(iv) In an isothermal process temperature remains constant.
(b) For adiabatic process use the relation
PV
γ
= P 1 V
γ
1 = const.
(1)
or
P =
P 1 V
γ
1
V γ
(2)
Work done on the gas
W = −
v 2
v 1
PdV = −
v 2
v 1
P 1 V
γ
1
dV
V γ = −P 1 V
γ
1
v 2
v 1
dV
V γ
P 1 V
γ
1
γ − 1
V
1−γ
2
− V
1−γ
1
=
1
γ − 1
P 1 V
γ
1 V
1−γ
2
− P 1 V 1
∴
W =
1
γ − 1
(P 2 V 2 − P 1 V 1 )
(3)
where we have used the relation P 1 V
γ
1 = P 2 V
γ
2 .
W is positive if V 1 > V 2 (compression) and negative if V 1 < V 2
(expansion).
10.42 Applying the gas equation
T A =
P A V A
n R
=
(1.013 × 10 5 )(44.8)
(2000)(8.31)
= 273 K