1. Xtra Edge February 2012 - Career Point

etween total mass and number of moles,
remember that if mtot is in kilograms, M must be
in kilograms per mole. The usual units for M are
grams per mole; be careful !
The internal energy change ∆U in any
thermodynamic process or series of processes in
independent of the path, whether the substance is
an ideal gas or not. This point is of the utmost
importance in the problems in this topic.
Sometimes you will be given enough information
about one path between the given initial and final
states to calculate ∆U for that path. Since ∆U is
the same for every possible path between the
same two states, you can then relate the various
energy quantities for other paths.
When a process consists of several distinct steps,
it often helps to make a chart showing Q, W, and
∆U for each step. Put these quantities for each
step on a different line, and arrange them so the
Q's, W's, and ∆U's form columns. Then you can
apply the first law to each line ; in addition, you
can add each column and apply the first law to the
sums. Do you see why ?
Using above steps, solve for the target variables.
Step 4: Evaluate your answer : Check your results for
reasonableness. In particular, make sure that each of
your answers has the correct algebraic sign.
Remember that a positive Q means that heat flows
into the system, and that a negative Q means that heat
flows into the system, and that a negative Q means
that heat flows out of the system. A positive W
means that work is done by the system on its
environment, while a negative W means that work is
done on the system by its environment.
Solved Examples
1. A metallic bob weighs 50 g in air. It it is immersed
in a liquid at a temperature of 25ºC, it weighs 45 g.
When the temperature of the liquid is raised to 100ºC,
it weighs 45.1 g. Calculate the coefficient of cubical
expansion of the liquid given that the coefficient of
linear expansion of the metal is 2 × 10 –6 (ºC) –1 .
Sol. Loss in weight in liquid at 25ºC = (50 – 45) = 5 gm
Weight of liquid displaced at 25ºC = V25ρ25g
∴ 5 = V25ρ25g ...(1)
Similarly, V100ρ100g = 50 – 45.1 = 4.9 gm ...(2)
From eq.(1) & (2) we get,
5 V25
ρ25
= .
4.
9 V100
ρ100
Now, V100 = V25(1 + γmetal × 75)= V25(1 + 3αmetal × 75)
= V25(1 + 3 × 12 × 10 –6 × 75)
or V100 = V25(1 + 0.0027) = V25 × 1.0027
Also, ρ25 = ρ100(1 + γ × 75)
where, γ = Required coefficient of expansion of the liquid
5
=
4.
9
V ρ ( 1 + 75γ)
1 ρ
25 ×
V25
× . 0027
100
or γ = 3.1 × 10 –4 (ºC) –1
XtraEdge for IIT-JEE 27 FEBRUARY 2012
100
1+ 75γ
=
1.
0027
2. A one litre flask contains some mercury. It is found
that at different temperature the volume of air inside
the flask remains the same. What is the volume of
mercury in flask ? Given that the coefficient of linear
expansion of glass = 9 × 10 –6 (ºC) –1 and coefficient of
volume expansion of mercury = 1.8 × 10 –4 (ºC –1 ).
Sol. Let V = Volume of the vessel
V' = Volume of mercury
For unoccupied volume to remain constant increase
in volume of mercury should be equal to increase in
volume of vessel.
∴ V' γm∆T = Vγg∆T
V × γ g
or V' =
γ m
∴
−6
1000×
27×
10
V' =
= 150 cm
−4
1.
8×
10
3
3. A clock with a metallic pendulum gains 6 seconds
each day when the temperature is 20ºC and loses 12
seconds each day when the temperature is 40ºC. Find
the coefficient of linear expansion of the metal.
Sol. Time taken for one oscillation of the pendulum is
T = 2 π
L
g
or T 2 = 4π 2 L
× .....(1)
g
Partially differentiating, we get
∆ L
.....(2)
2T∆t = 4π 2 × g
Dividing (2) by (1), we get
∆ T ∆ L α L∆t
1
= = = α∆t
T 2L
2L
2
where ∆t is the change in temperature. Now,
One day = 24 hours = 86400 sec
Let t be the temperature at which the clock keeps
correct time.
At 20ºC, the gain in time is
1
6 = α × (t – 20) × 86400 ....(3)
2
At 40ºC, the loss in time is
1
12 = α× (40 – t) × 86400 ...(4)
2
Dividing (4) by (3), we have
12 40 − t
=
6 t − 20
80
which gives t = ºC.
3
Using the value in equation(3), we have
1 ⎛ 80 ⎞
6 = × α × ⎜ − 20⎟⎠
× 86400
2 ⎝ 3
which gives α = 2.1 × 10 –5 perºC