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Specific Heat Capacity Of Gases

Specific Heat Capacity Of Gases

Specific heat of a gas is numerically equal to the amount of heat necessary to raise the temperature of unit mass of gas by 1°C. In order to raise the temperature of unit mass of a gas through 1°C more heat will be required if the gas was kept at constant pressure than when it is at constant volume.

(i) Molar Specific heat capacity at constant Volume: The amount of heat required to raise the temperature of 1 mole of gas by 1 °C at constant volume is called the molar specific heat and it is represented by C_v.

            C_v = (ΔQ/mΔt)_v = constant

By first law of thermodynamics ΔQ = ΔU + W

But W = 0 for isochoric process, then ΔQ = ΔU

by definition of specific heat ΔQ = nC_vΔT

Where CV is the specific heat (for 1 mole of gas), then ΔU = nCvΔT The relation ΔU = nCvΔT, is used to find the change in internal energy of the system and is valid for any process where a temperature change has taken place. Consider two isotherms on the P-V diagram: Process 1 –> 2 represents an isochoric process Process 1 –> 3 represents an isobaric process

In both processes the temperature has changes from T₁ to T₂ as 2 and 3 lie on the same isotherm. Thus change in internal energ

             U_1–>2 = U_1–>3 = nC_vΔT

(ii) Molar Specific heat capacity at constant Pressure: The amount of heat required to raise the temperature of 1 mol of gas by 1°C keeping its pressure constant, is called molar specific heat at constant pressure and it is represented by C_p,

             C_p = (ΔQ/mΔT)_P  = constant

(iii) Relation between C_p and C_v: For an isobaric process by definition

From first law ΔQ = ΔU + W

=> nC_PΔT = nC_VΔT + W      [∴ ΔU = nC_VΔT for any process]

For isobaric process

W = PΔV = nRΔT            nC_PΔT = nC_VΔT + nRΔT

=> C_P = C_V + R      or     C_P ­- C_V = R     (Meyer’s relation)

(iv) Relation between specific heat (CP and CV) and degrees of freedom: If f is the number of degree of freedom of a gas molecule then the internal energy of n moles of that gas is given by

U = f × 1/2 nRT            (from law of equipartition of energy)

(∴ U = f/2 kT = f/2 R/N T (for n moles) = nN × f/2 R/N T = n f/2 RT)

=> dU = f × 1/2 nRdT

Also dU = nCVdT, 

So, nCVdT = f × 1/2 nRdTf/2 R

but CP = CV + R

∴ C_P = (f/2 f/2 + 1)R. This is the relation between specific heat ratio and degree of freedom.

(v) Adiabatic Expansion of an Ideal gas: As defined earlier in adiabatic process ΔQ = 0

For a system containing an ideal gas adiabatic process can happen in two ways

(i) if the system boundary is adiabatic

(ii) if the system boundary is diathermic but the process takes place so fast that their is no time for the transfer of the heat. For example, Propagation of sound in air.  

For an adiabatic process as shall be proved

PV? = a constant, here g = C_P/ C_V. ? is known as adiabatic constant

Since PV = nRT for an ideal gas

P = nRT/V so,

(nRT/V)V^g = Constant           ∴ TV^g–1 = a constant

Proof: Let n moles of an ideal gas expand adiabatically by a small amount ΔV.

By first law   ΔQ = ΔU + PΔV

As ΔQ = 0    we have ΔU = –PΔV

But       U = nC_vΔT

So,       nΔT = –(P/C_v)ΔV                                                                               …(i)

From the ideal gas law (after differentiating) ,          PΔV = VΔP = nRΔT

Replacing R by its equal, C_P ­- C_V leads to this,         nΔT = PΔV + VΔP/C_p – C_v        …(ii)

equating (1) and (2) with a little algebra leads to

            ΔP/P + (C_p/C_v) ΔV/V = 0            Cp/Cv = g

and integrating we have, ln P + g In V = a constant,

or PV^g = a constant, hence proved.

Again g = C_p/C_v = C_v + R/C_V = 1 + R/C_v = 1 + R/(1/2 fR) = 1 + 2/f

(vi) Work done in an adiabatic process (P₁V₁T₁) to (P₂V₂T₂)

For adiabatic process, W = –ΔU = –n C_vΔT

From ideal gas equation P₁V₁ – P₂V₂ = nR (T₁ – T₂) = – nRT

     W = C_v(P₁V₁ – P₂V₂)/R(P₁V₁ – P₂V₂)/?–1 = nR(T₁ – T₂)/–1

(vii) The table shows the values of f, Cv, Cp and g for different gasses:

Nature of gas	Degree of freedom f = (T+R+V)	Cv	Cp	g
Monatomic	3+0+0 = 3	3	5	5/3
Diatomic	3+2+0 = 5	5	7	7/5
Polyatomic (linear)	3+2+0 = 5	5	7	7/5
Polyatomic (non-linear)	3+2+1 = 6	3R	4R	4/3

Note: At room temperature the energy associated with vibrational motion is negligible in comparison to translational and rotational KE.

Expressions For ΔU, W, And ΔQ For Different Process

Processes	Relation between thermodynamic variables	Work Done (W)	Heat Exchange (ΔQ)
Isothermal Process (T constant)	P∝1/V	W = 2.303 nRT log10 (V2/V1)	ΔQ = 2.303 nRT log10(V2/V1)
Adiabatic Process (No heat exchange)	PVg = constant	W = (P1V1 – P2V2)/(g–1) = nR(T1– T2)/(g-1)	ΔQ = 0
Isochoric   Process (V = constant)	P ∝T	W = 0	ΔQ = n CV ΔT* (Use definition of Cv)
Isobaric Process (P = Constant)	V ∝T	W = P Δ V = P(V2 –V2) W = nR (T2 –T1)	ΔQ = n CP ΔT* (Use definition of Cp)

Proof of Bernoulli Equation

The Bernoulli Equation for an incompressible, steady Fluid Flow:

The Bernoulli Equation is a statement derived from conservation of energy and work-energy ideas that come from Newton's Laws of Motion.

Statement:It states that the total energy (pressure energy, potential energy and kinetic energy) of an incompressible and non–viscous fluid in steady flow through a pipe remains constant throughout the flow, provided there is no source or sink of the fluid along the length of the pipe.

This statement is based on the assumption that there is no loss of energy due to friction.

To prove Bernoulli’s theorem, we make the following assumptions:

1. The liquid is incompressible.   

2. The liquid is non–viscous. 

3. The flow is steady and the velocity of the liquid is less than the critical velocity for the liquid.

Proof of Bernoulli’s Theorem:
Imagine an incompressible and non–viscous liquid to be flowing through a pipe of varying cross–sectional area as shown in Fig. The liquid enters the pipe with a normal velocity v₁ and at a height h₁ above the reference level (earth’s surface). It leaves the pipe with a normal velocity v₂ at the narrow end B of cross–sectional area a₂ and at a height h₂ above the earth’s surface.

We examine a fluid section of mass m traveling to the right as shown in the schematic above. The net work done in moving the fluid is Eq.(1) where F denotes a force and an x a displacement. The second term picked up its negative sign because the force and displacement are in opposite directions. Pressure is the force exerted over the cross-sectional area, or P = F/A. Rewriting this as F = PA and substituting into Eq.(1) we find that Eq.(2)

The displaced fluid volume V is the cross-sectional area A times the thickness x. This volume remains constant for an incompressible fluid, so Eq.(3) Using Eq.(3) in Eq.(2) we have Eq.(4) Since work has been done, there has been a change in the mechanical energy of the fluid segment. This energy change is found with the help of the next diagram. The energy change between the initial and final positions is given by Eq.(5) Here, the the kinetic energy K = mv²/2 where m is the fluid mass and v is the speed of the fluid. The potential energy U = mgh where g is the acceleration of gravity, andh is average fluid height. The work-energy theorem says that the net work done is equal to the change in the system energy. This can be written as Eq.(6) Substitution of Eq.(4) and Eq.(5) into Eq.(6) yields Eq.(7) Dividing Eq.(7) by the fluid volume, V gives us Eq.(8)     where Eq.(9) is the fluid mass density. To complete our derivation, we reorganize Eq.(8). Eq.(10) Finally, note that Eq.(10) is true for any two positions. Therefore, Eq.(11) Equation (11) is commonly referred to as Bernoulli's equation. Keep in mind that this expression was restricted to incompressible fluids and smooth fluid flows.

Work, Power & Energy

Work

Work is said to be done when a force applied on the body displaces the body through a certain distance in the direction of force.

Let a constant force  be applied on the body such that it makes an angle θ with the horizontal and body is displaced through a distance s

By resolving force  into two components :

(i) F cos θ   in the direction of displacement of the body.

(ii) F sin θ  in the perpendicular direction of displacement of the body.

Since body is being displaced in the direction of , therefore work done by the force in displacing the body through a distance s is given by

                

or             

Thus work done by a force is equal to the scalar or dot product of the force and the displacement of the body.

If a number of force  are acting on a body and it shifts from position vector  to position vector  then .

Unit of Work

Since work is the product of force and distance, unit of work is unit of force and unit of distance. If force is in N and distance is in metres, unit of work will be N-m. In SI system of unit, 1N-m = 1 J. The term Joule s used for one N-m work done, and letter J is used as a symbol of Joule. Hence, one Joule may be defined as the amount of work done by one Newton force when it moves one metre distance in its direction.

Energy

The energy of a body is defined as its capacity for doing work. Since energy of a body is the total quantity of work done therefore it is a scalar quantity.

Power

Power of a body is defined as the rate at which the body can do the work.

Average power 

Instantaneous power          [As ]

                                                           [As ]

i.e. power is equal to the scalar product of force with velocity.

If work done by the two bodies is same then power 

i.e. the body which perform the given work in lesser time possess more power and vice-versa. As power = work/time, any unit of power multiplied by a unit of time gives unit of work (or energy) and not power.

Work Done by Variable Force

Let a varying force P move a body by distance s as shown by force-displacement curve in Fig.. At an instance, when force P is acting and the body moves by an elemental distance ' δs', then work done at that instance

= P δ s

Hence work done by the varying force in moving the body by distance ‘s’

= Σ P δ s

and it is equal to the area under the force-displacement curve. The variation of force is in a regular fashion so that its value at any instance can be represented by an expression. Then the work done can be evaluated by suitable integration. In such a case,

 work done= ΣPs = ∫ Pds

Alternating Current

Alternating Current

The basic principle of the ac generator is a direct consequence of Faraday's law of induction. When a conducting loop is rotated in a magnetic field at constant angular frequency ω a sinusoidal voltage (emf) is induced in the loop. This instantaneous voltage is,

            V = V₀ sin ωt                                       ... (i)

The usual circuit diagram symbol for an ac source is shown in Figure.

In Equation (i) V₀ is the maximum output voltage of the ac generator, or the voltage amplitude and w is the angular frequency, equal to 2p times the frequency f.

            ω = 2pf

The frequency of ac in India is 50 Hz, i.e.,

            f = 50 Hz

So,       ω = 2pf » 314 rad/s

The time of one cycle is known as time period T, the number of cycles per second the frequency f.

            T = 1/f     or  T = 2π/ω

A sinusoidal current might be described as,

            i = i₀ sin ωt

If an alternating current is passed through an ordinary ammeter or voltmeter, it will record the mean value for the complete cycle, as the quantity to be measured varies with time. The average value of current for one cycle is,

Thus,      _{one cycle} = 0

Similarly, the average value of the voltage (or emf) for one cycle is zero.

              _{one cycle} = 0

Since, these averages for the whole cycle are zero, the dc instrument will indicate zero deflection. In ac, the average value of current is defined as its average taken over half the cycle. Hence,

This is sometimes simply written as, i_av. Hence,

               i_av = _{Half Cycle} = 2/π i₀ ≈ 0.637 i0

Similarly, V_av =  2/π V₀ ≈ 0.637 V₀

A dc meter can be used in an ac circuit if it is connected in the full wave rectifier circuit. The average value of the rectified current is the same as the average current in any half cycle, i.e., 2/π time the maximum current i₀. A more useful way to describe a quantity is the root mean square (rms) value. We square the instantaneous current, take the average (mean) value of i² and finally take the square root of that average. This procedure defines the root-mean-square current denoted as irms. Even when i is negative, i² is always positive so irms is never zero (unless i is zero at every instant). Hence,

Similarly, we get RMS value V_rms = V₀/√2 ≈ 0.707 V₀

The square root of the mean square value is called the virtual value and is the value give by ac instruments.

Thus, when we speak of our house hold power supply as 220 volts ac, this means that the rms voltage is 220 volts and its voltage amplitude is,

            V₀ = √2V_rms = 311 volt

Form Factor

The ratio,       

rms value/average value = (V₀/√2)/(2V₀/π)

                                   = π/2√2 = 1.111

is known as or factor,

Note:

1. The average value of sin wt, cos wt, sin2wt, cos2wt, etc. is zero because it is positive half of the time and negative rest half of the time. Thus,

2. The average value of sin² wt and cos² wt is 1/2

3. Like SHM, general expressions of current/voltage in an sinusoidal ac are,

     = i₀ sin (ωt ± f),                   V = V₀ sin (ωt ± f)

or i = i₀ cos (ωt ± f),        and     V = V₀ cos (ωt ± f)

Ray Optics - Optical Instruments

Optical Instruments

Optical instruments are devices that are used to processes light waves to enhance an image for viewing, or analyze light waves to determine one of a number of characteristic properties. The first optical instrument invented was telescope and was used for magnification of distant images, and microscopes used for magnifying very tiny images. 

Simple Microscope

It is an optical instrument used to see very small objects. It's magnifying power is given by

m = Visual angle with instrument (β)/Visual angle when object is placed at least distance of distinct vision(α)

Simple microscope

(i) It is a single convex lens of lesser focal length.

(ii) Also called magnifying glass or reading lens.

(iii) Magnification's, when final image is formed at D and ∞

     (i.e. m_D and m_∞)    m_D = (1 + D/f)_max

     and       m_∞ = (D/f)_min

(iv) If lens is kept at a distance a from the eye then m_D = 1 + D – a /f and m_∞ = D–a/f

Compound Microscope

(i) Consist of two converging lenses called objective and eye lens.

(ii) f_{eye lens} > f_objective and (diameter)_{eye lens} > (diameter)_objective

(iii) Intermediate image is real and enlarged.

(iv) Final image is magnified, virtual and inverted.

(v) u_o = Distance of object from objective(o), v_o = Distance of image (A'B') formed by objective from objective, u_e = Distance of A'B' from eye lens, v_e = Distance of final image from eye lens, f_o = Focal length of objective, f_e = Focal length of eye lens.

(vi) Final image is formed at D: Magnification m_D = – v_o/u_o (1+ D/f_e) and length of the microscope tube (distance between two lenses) is L_D = v_o + u_e.

Generally object is placed very near to the principal focus of the objective hence u_o = f_o. The eye piece is also of small focal length and the image formed by the objective is also near to the eye piece.

So v_o = L_D, the length of the tube.

Hence, we can write m_D = –L/d_o (1+ D/f_e)

(vii) Final image is formed at ∞: magnification m∞ = –v_o/u_o.D/f_e and length of tube L∞ = v_o + f_e

      In terms of length m∞ = (L_∞ – f_o – f_e)D/f_o f_e

(viii) For large magnification of the compound microscope, both f_o and f_e should be small.

(ix) If the length of the tube of microscope increases, then its magnifying power increases.

(x) The magnifying power of the compound microscope may be expressed as M = m_o × m_e; where m_o is the magnification of the objective and m_e is magnifying power of eye piece.

Astronomical Telescope

By astronomical telescope heavenly bodies are seen.

(i) f_objective >; f_{eye lens} and d_objective >; d_{eye lens}

(ii) Intermediate image is real, inverted and small.

(iii) Final image is virtual, inverted and small.

(iv) Magnification: m_D = –f_o/f_e (1+f_e/D) and m_∞ = –f_o/f_e

(v) Length : L_D = f_o+ u_e and L_∞ = f_o + f_e

Resolving Limit and Resolving Power

Microscope: In reference to a microscope, the minimum distance between two lines at which they are just distinct is called Resolving limit (RL) and it's reciprocal is called Resolving power (RP)

λ = Wavelength of light used to illuminate the object,

µ = Refractive index of the medium between object and objective,

θ = Half angle of the cone of light from the point object,

µ sin θ = Numerical aperture.

Telescope: Smallest angular separations (dθ) between two distant objects, whose images are separated in the telescope is calledresolving limit. So revolving limit dθ = 1.22λ/a and resolving power  where a = aperture of objective.

Refraction of Light through Plane Surfaces

Refraction of Light

The bending of the ray of light passing from one medium to the other medium is called refraction.

(i) The refraction of light takes place on going from one medium to another because the speed of light is different in the two media.

(ii) Greater the difference in the speeds of light in the two media, greater will be the amount of refraction.

(iii) A medium in which the speed of light is more is known as optically rarer medium and a medium is which the speed of light is less, is known as optically denser medium.

(iv) When a ray of light goes from a rarer medium to a denser medium, it bends towards the normal.

(v) When a ray of light goes from a denser medium to a rarer medium, it bends away from the normal.

Refractive Index

(i) Refractive index of a medium is that characteristic which decides speed of light in it.

(ii) It is a scalar, unit less and dimensionless quantity.

(iii) Absolute refractive index:

When light travels from vacuum to any transparent medium then refractive index of medium w.r.t. vacuum is called it's absolute refractive index i.e. _vacuumµ_medium = c/v

Absolute refractive indices for glass, water and diamond are respectively µ_g = 3/2 = 1.5, µ_w = 4/3 = 1.33 and µ_D = 12/5 = 2.4

(iv) Relative refractive index:

When light travels from medium (1) to medium (2) then refractive index of medium (2) w.r.t. medium (1) is called it's relative refractive index i.e. ₁µ₂ = µ₂/µ₁ = v₁/v₂ (where v₁ and v₂ are the speed of light in medium 1 and 2 respectively).

(v) When we say refractive index we mean absolute refractive index.

(vi) The minimum value of absolute refractive index is 1. For air it is very near to 1. ( 1.003)

(vii) Cauchy's equation : µ = A + B/λ² + C/λ⁴ + ...         (1_Red > 1_Violet so m_Red < m_Violet)

(viii) If a light ray travels from medium (1) to medium (2), then ₁µ ₂ =  µ₂/µ₁ =  λ₁/ λ₂ = v₁/v₂

(ix) Dependence of Refractive index:

(a) Nature of the media of incidence and refraction.

(b) Colour of light or wavelength of light.

(c) Temperature of the media : Refractive index decreases with the increase in temperature.

(x) Reversibility of light and refraction through several media:

(A) ₁µ₂ = 1/₂µ₁      (B)   ₁µ₂x₂µ₃x₃µ₁ = 1 or ₂µ₃ = ₁µ₃/₁µ2

Snell's law:

The ratio of sine of the angle of incidence to the angle of refraction (r) is a constant called refractive index

i.e.   sin i/sin r = µ (a constant).

For two media, Snell's law an be written as ₁µ₂ = µ₂/µ₁ = sin i/sin r or µ₁sin i = µ₂sin r

Refraction Through Glass

Lateral Shift:

The refracting surfaces of a glass slab are parallel to each other. When a light ray passes through a glass slab it is refracted twice at the two parallel faces and finally emerges out parallel to it's incident direction i.e. the ray undergoes no deviation δ = 0 . The angle of emergence (e) is equal to the angle of incidence (i)

The Lateral shift of the ray is the perpendicular distance between the incident and the emergent ray, and it is given by MN = t secrsin (i – r)

Apparent Depth

When object is in denser medium and observer is in rarer medium

(i) µ = Real depth/Apparent depth = h/h'

(ii) Real > Apparent depth

(iii) Shift d = h – h' = (1 – 1/µ)h.

(iv) If a beaker contains various immiscible liquids as shown then Apparent depth of

bottom d₁/µ₁ + d₂/µ₂ + d₃/µ₃ + ...

Object is in rarer medium and observer is in denser medium

(i) µ = h'/h

(ii) Real depth < Apparent depth.

(iii) d = (µ = 1)h

Total Internal Reflection (TIR)

When a ray of light goes from denser to rarer medium it bends away from the normal and as the angle of incidence in denser medium increases, the angle of refraction in rarer medium also increases and at a certain angle, angle of refraction becomes 90°, this angle of incidence is called critical angle (C).

When Angle of incidence exceeds the critical angle than light ray comes back in to the same medium after reflection from interface. This phenomenon is called Total internal reflection (TIR).

(i) µ = 1/sin C = cosec C where µ means; _Raserµ_Denser

(ii) Conditions for Total Internal Reflection

(a) The ray must travel from denser medium to rarer medium.

(b) The angle of incidence i must be greater than critical angle C