EP2200: THERMODYNAMICS: ASSIGNMENT 1
1. We have an equation of the following form:
f (x, y, z) = 0
(1)
Using this we can write: x = x(y, z), y = y(x, z), z = z(x, y), i.e, x as a function of y
and z and so on. Thus, we have
∂x
∂x
dy +
dz
(2)
dx =
∂y z
∂z y
∂y
∂y
dy =
dx +
dz
(3)
∂x z
∂z x
∂z
∂z
dz =
dx +
dy
(4)
∂x y
∂y x
Prove that:
(a)
∂x
∂y
= 1;
∂y z ∂x z
∂x
∂x
∂y
=−
∂y z ∂z x
∂z y
3 marks
(b)
∂y
∂z
= 1;
∂z x ∂y x
∂z
∂y
∂y
=−
∂z x ∂x y
∂x z
3 marks
(c)
∂x
∂z
∂x
∂z
y
y
∂z
∂x
∂z
∂y
= 1;
y
=−
x
∂x
∂y
z
3 marks
2. Suppose we are given,
dz = f1 (x, y)dx + f2 (x, y)dy.
Prove that z = z(x, y), i.e, z is a function of x and y, only if
∂f1
∂f2
=
∂y x
∂x y
3 marks
3. For a stretched wire, we have equation of state f (T , L, T ) = 0, where T → tension,
T → temperature, L → length.
Other properties of the 1wire:
cross-sectional area A,
L ∂T
∂L
Young’s modulus Y = A ∂L T , linear expansivity α = L ∂T T . Prove that
(a)
dL = αL dT +
L
dT.
αY
5 marks
(b)
dT = −AαY dT +
A
Y dL.
L
5 marks
4. (a) Using the same notations as in Problem 3, write down the expression for infinitesimal amount of work done on changing length the stretched wire by dL. Prove
that, W 6= W (L, T ), i.e, total work cannot be written as function of L and T ,
unless T is independent of T .
2 marks
(b) The proof in part (a) means infinitesimal amount of work is an inexact differential
dW
¯ . Starting from this, mathematically define:
– dU → change in internal energy
– dQ
¯ → heat absorbed,
such that, dW
¯ = dU − dQ,
¯ noting that dU is an exact differential, dQ
¯ is an inexact
differential. These definitions should be in terms of two unknown functions of L
and T . Call them g1 (L, T ), g2 (L, T ). What is the partial differential equation
that relates g1 (L, T ) and g2 (L, T )?
3 marks
dQ
¯
(c) Find g1 (L, T ) and g2 (L, T ) in terms of CL , CT and α, where CL = dT L , CT =
dQ
¯
, and α is as defined in Problem 3.
dT T
5 marks
5. Prove that for the stretched wire
(a)
∂U
∂T
T
= CT + T Lα.
10 marks
(b)
∂U
∂T
=
T
CL − CT
LT
+
.
AY α
AY
8 marks