1. Xtra Edge February 2012 - Career Point

(f) ∆U = ncv∆T where cv =
piVi
− p f V f
(g) W =
γ −1
(h) Graphs
P
P
=
R
γ −1
nR(
Ti
− T f )
γ −1
V
T
Please note that P-V graph line (isotherm) is
steeper.
For isochoric process
(a) P ∝ T
(b) W = 0
(c) q = ∆U
(d) ∆U = nCv∆T
(e) Graphs
R
where Cv =
γ −1
P
P
V
V
For isobaric process
(a) V ∝ T
T
(b) W = P∆V = P(Vf – Vi) = nR(Tf – Ti)
(c) ∆U = nCv∆T
(d) q = nCp∆T
(e) Graphs
P
P
V
V
T
T
For a cyclic process
(a) ∆U = 0 ⇒ q = W
(b) Work done is the area enclosed in p-V graph.
For any process depicted by P-V diagram, area under
the graph represents the word done.
Kirchoff's law states that good absorbers are good
emitters also.
Problem solving Strategy : Thermal Expansion
Step 1: Identify the relevant concepts: Decide
whether the problem involves changes in length
(linear thermal expansion) or in volume (volume
thermal expansion)
Step 2: Set up the problem using the following steps:
Eq. ∆L = αL0∆T for linear expansion and
Eq. ∆V = βV0∆T for volume expansion.
Identify which quantities in Eq. ∆L = αL0∆T or
∆V = βV0∆T are known and which are the
unknown target variables.
V
T
T
Step 3: Execute the solution as follows:
Solve for the target variables. Often you will be
given two temperatures and asked to compute ∆T.
Or you may be given an initial temperature T0 and
asked to find a final temperature corresponding to
a given length or volume change. In this case,
plan to find ∆T first; then the final temperature is
T0 + ∆T.
Unit consistency is crucial, as always. L0 and ∆L
(or V0 ∆V) must have the same units, and if you
use a value or α or β in K –1 or (Cº) –1 , then ∆T
must be in kelvins or Celsius degrees (Cº). But
you can use K and Cº interchangeably.
Step 4: Evaluate your answer: Check whether your
results make sense. Remember that the sizes of holes
in a material expand with temperature just as the
same way as any other linear dimension, and the
volume of a hole (such as the volume of a container)
expands the same way as the corresponding solid
shape.
Problem solving strategy : Thermodynamics I st Law
Step 1: Identify the relevant concepts : The first law
of thermodynamics is the statement of the law of
conservation of energy in its most general form. You
can apply it to any situation in which you are
concerned with changes in the internal energy of a
system, with heat flow into or out of a system, and/or
with work done by or on a system.
Step 2: Set up the problem using the following steps
Carefully define what the thermodynamics system is.
The first law of thermodynamics focuses on
systems that go through thermodynamic
processes. Some problems involve processes
with more than one step. so make sure that you
identify the initial and final state for each step.
Identify the known quantities and the target
variables.
Check whether you have enough equations. The
first law, ∆U = Q – W, can be applied just once to
each step in a thermodynamic process, so you will
often need additional equations. These often
include Eq. W = ∫ pdV
for the work done in a
XtraEdge for IIT-JEE 26 FEBRUARY 2012
V2
V1
volume change and the equation of state of the
material that makes up the thermodynamic system
(for an ideal gas, pV = nRT).
Step 3: Execute the solution as follows :
You shouldn't be surprised to be told that
consistent units are essential. If p is a Pa and V in
m 3 , then W is in joules. Otherwise, you may want
to convert the pressure and volume units into
units of Pa and m 3 . If a heat capacity is given in
terms of calories, usually the simplest procedure
is to convert it to joules. Be especially careful
with moles. When you use n = mtot/M to convert