E/S

.(a) When the temperature of a substance is increased, it expands. The heat energy which is supplied to the substance is gained by the constituent particles of the substance as its kinetic energy.

Because of this the collisions between the constituents particles are accompanied with greater force which increase the distance between the constituent particles.

∆l = lα∆T ; ∆A = Aβ∆T ; ∆V = Vγ∆T or l' = l (1 + α∆T) ; A' = A(1 + β∆T) ;

V' = V(1 + γ∆T)

(b) Also ρ = ρ'(1 + γ∆T) where ρ' is the density at higher temperature clearly ρ' < ρ for substances which have positive value of γ

* β = 2α and γ = 3α

Water has negative value of γ for certain temperature range (0º to 4ºC). This means that for that temperature range the volume decreases with increase in temperature. In other words the density increases with increase in temperature.

0 ml 5 ml 10 ml 15 ml 20 ml 25 ml 30 ml

If a liquid is kept in a container and the temperature of the system is increased then the volume of the liquid as well as the container increases. The apparent change in volume of the liquid as shown by the scale is

∆Vapp = V(γ – 3α) ∆T

Where V is the volume of liquid at lower temperature

∆V_app is the apparent change in volume

γ is the coefficient of cubical expansion of liquid α is the coefficients of linear expansion of the container.

Loss or gain in time by a pendulum clock with change in temperature is ∆t =

2

1α(∆T) × t

Where ∆t is the loss or gain in time in a time interval t

∆T is change in temperature and d is coefficient of linear expansion.

If a rod is heated or cooled but not allowed to expand or contract then the thermal stresses developed

A

F = γα∆T.

If a scale is calibrated at a temperature T1 but used at a temperature T2, then the observed reading will be wrong. In this case the actual reading is given by R = R0(1 + α∆T)

Where R0 is the observed reading, R is the actual reading.

For difference between two rods to the same at all temperatures l 1α1 = l2α2.

Thermodynamics

According to first law of thermodynamics q = ∆U + W

For an isothermal process (for a gaseous system) (a) The pressure volume relationship is ρV = constt.

(b) ∆U = 0 (c) q = W

(d) W = 2.303 nRT log10

i f V

V = 2.303 nRT log10

f i p p

(e) Graphs T2 > T1

T2

T1

P

V P

T V

T These lines are called isotherms (parameters at constant temperature)

For an adiabatic process (for a gaseous system) (a) The pressure-volume relationship is PV^γ = constt.

(b) The pressure-volume-temperature relationship is T

PV = constt.

(c) From (a) and (b) TV^γ–I = constt.

(d) q = 0 (e) W = –∆U

Thermal Expansion, Thermodynamics

P ^HYSICS F UNDAMENTAL F ^OR IIT-J ^EE

KEY CONCEPTS & PROBLEM SOLVING STRATEGY

(f) ∆U = ncv∆T where cv = 1 R

− γ

(g) W =

1 V p V p_{i i f f}

− γ

− =

1 ) T T ( nR _{i f}

− γ

−

(h) Graphs

^V

P

P

T 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 = nC_v∆T where Cv = 1 R

− γ (e) Graphs

P

V P

T V

T For isobaric process

(a) V ∝ T

(b) W = P∆V = P(V_f – Vi) = nR(Tf – Ti) (c) ∆U = nC_v∆T

(d) q = nCp∆T (e) Graphs

P

V P

T V

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.

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. ⁼

∫

²

1 V

V dV p

W for the work done in a 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 m³, then W is in joules. Otherwise, you may want to convert the pressure and volume units into units of Pa and m³. 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

between 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.

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,

9 . 4

5 =

100 25 100

25 . V

V ρ

ρ

Now, V100 = V25(1 + γ_metal × 75)= V25(1 + 3α_metal × 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

4.9 5 =

100 100 25

25 (1 75 )

0027 . 1 V

V

ρ γ +

×ρ

× =

0027 . 1

75 1+ γ

or γ = 3.1 × 10^–4 (ºC)^–1

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 or V' =

m

V g

γ γ

×

∴ V' = ₄ ⁶

10 8 . 1

10 27 1000

−

−

×

×

× = 150 cm³

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 = g

2π L or T² = 4π² × g

L ...(1) Partially differentiating, we get

2T∆t = 4π² × g

∆L

...(2) Dividing (2) by (1), we get

T

∆T

= 2L

∆L

= 2L L∆t

α = t

21 ∆α 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

6 =

2

1α × (t – 20) × 86400 ....(3) At 40ºC, the loss in time is

12 = 2

1α× (40 – t) × 86400 ...(4) Dividing (4) by (3), we have

6 12 =

20 t

t 40

−

−

which gives t = 3 80ºC.

Using the value in equation(3), we have 6 = 2

1 × α × 



 



 − 20 3

80 × 86400 which gives α = 2.1 × 10^–5 perºC

Solved Examples

4. A piston can freely move inside a horizontal cylinder closed from both ends. Initially, the piston separates the inside space of the cylinder into two equal parts each of volume V0, in which an ideal gas is contained under the same pressure p0 and at the same temperature. What work has to be performed in order to increase isothermally the volume of one part of gas η times compared to that of the other by slowly moving the piston ?

Sol. Let volume of chambers changes by ∆V. According to the problem, the final volume of left chamber is η times final volume of right chamber.

∴ V0 + ∆V = η(V0 – ∆V)

or ∆V = V₀

1 1

 



 + η

− η

P0,v0,T0

P0,v0,T0

As piston is moved slowly therefore, change in kinetic energy is zero. By work-energy theorem, we can write

Wgas in right chamber + Wgas in left chamber + W_Agent^ext = ∆KE

Agentext

W = (Wgas(R) + Wgas(L))

We know that in isothermal process, work done is given by

W = nRT ln _



 





i f

V V

∴ Work done by gas in left chamber (W_L) = P0V0 ln _



 



 +∆

0 0

V V

V = P0V0 ln _



 



 + η

η 1 2

Similarly, work done by gas in right chamber (WR) = P0V0 ln _



 



 −∆

0 0

V V

V = P0V0 ln _



 



 + η

η 1 2

Agentext

W = –P0V0 ln _



 



 + η

η 1

2 – P0V0 ln _



 



 + η

η 1 2

= P0V0 ln

2

4 1

 



 η + η

5. A smooth vertical tube having two different sections is open from both ends equipped with two pistons of different areas figure. Each piston slides within a respective tube section. One mole of ideal gas is enclosed between the pistons tied with a non-stretchable thread. The cross-sectional area of the upper piston is ∆S greater than that of the lower one.

The combined mass of the two pistons is equal to m.

The outside air pressure is P0. By how many kelvins must the gas between the pistons be heated to shift the pistons through l.

P0

P0

Sol. Let A1 = Cross section of upper piston A2 = Cross section of lower piston T = Tension in the string

P = Gas pressure

m1 = Mass of upper piston m2 = Mass of lower piston Now, consider FBD of upper piston

PA1 m1g P0 A1

From equilibrium consideration of upper piston we get, P0A1 + T + m1g = PA1

Similarly, consider FBD of lower piston PA2

m2g P0 A2

T

∴ P0A2 + T = m2g + PA2

Eliminating T, we get P = P0 +

2 1

2 1

A A

g ) m m (

− +

According to problem m = m1 + m2

and ∆S = A₁ – A2

∴ P = P0 + S mg

∆ Now, PV = RT

or P∆V = R∆T or ∆T = R P∆V But ∆V = (A₁ – A2)l = ∆S. l

∴ ∆T = 



 



 + ∆

S P₀ mg ∆S.l

l l

l

Qualitative Analysis :

Qualitative analysis of an organic compound involves the detection of various elements present in it. The elements commonly present in organic compounds are carbon, hydrogen, oxygen, nitrogen, halogens, sulphur and sometimes phosphorus.

Detection of Carbon and Hydrogen :

This is done by heating the given organic compound with dry cupric oxide in a hard glass test tube when carbon present is oxidised to carbon dioxide and hydrogen is oxidised to water.

C + 2 CuO →^∆ CO2 + 2Cu 2H + CuO →^∆ H2O + Cu

Carbon dioxide turns lime water milky.

Ca(OH)2 +

)

In document PROYECTO FIN DE CARRERA (página 63-68)