Thermodynamic Change or Thermodynamic Process

Thermodynamic state of a system is characterized by its parameters P,V and T. A change in one or more parameters results in a change in the state of the system.

A process by which one or more parameters of thermodynamic system undergo a change is called a thermodynamic process or a thermodynamic change. Following are the various types of thermodynamic process.

(a) Isothermal Process

The process in which change in pressure and volume takes place at a constant temperature, is called a isothermal change. It may be noted that in such a change total amount of heat of the system does not remain constant.

Consider some gas contained in a barrel of conducting material, fitted with an airtight piston. As the gas is compressed as shown in the below figure, heat is liberated.

The liberated heat moves out to keep the temperature constant. When the gas is expanded as shown in the below figure, heat is absorbed. This absorption is compensated by the transfer of heat from the surroundings, again keeping the temperature constant. Such a change is called an isothermal change. P-V diagram for an isothermal change is shown in below figure. The curve is called an ‘isothermal’.

It may be noted that in such a change total amount of heat of the system does not remain constant. 

Since temperature remains constant, Boyle’s law can be applied to study this change.

Therefore, the equation of state for an isothermal change is given by,

PV=Constant

If P₁, V₁, and P₂, V₂ are the parameters of system in two different states, then for an isothermal change,

P₁V₁ = P₂V₂ = Constant 

(b) Adiabatic Process

The process in which change in pressure and volume and temperature takes place without any heat entering or leaving the system is called adiabatic change.

Consider some gas contained in an insulating barrel fitted with an air tight piston. On compressing the gas, heat is liberated. This liberated heat cannot move out due to the insulation of the barrel, resulting in a rise of temperature. On expanding the gas, heat is absorbed. This results in a fall of temperature. P-V diagram for an adiabatic change is shown in below figure.

It may be noted that the total heat of the system, undergoing an adiabatic change always remains constant.

The equation of state for adiabatic change is given by,

Where ‘’ is the ratio between the two specific heats of the gas.

Here C_p and C_v are the molar specific heat capacities of the gas at constant pressure and at constant volume respectively.

For the diatomic gases at medium temperatures (250 K to 750 K), ‘’ has a value of 1.42.

(c) Isobaric Process

The process in which change in volume and temperature of a gas take place at a constant pressure is called an isobaric process.   

Consider some gas contained in a barrel. Supply some quantity of heat to increase its temperature from T₁ to T₂ without changing its pressure ‘P’. It will be observed that the gas expand. The change is said to be isobaric change. P-V diagram for an isobaric change is a line parallel to V-axis as shown in below figure.

(d) Isochoric Process

The process in which changes in pressure and temperature take place in such a way that the volume of the system remains constant, is called isochoric process.

Supply a quantity of heat Q to a gas contained in a barrel. The gas tends to expand. Let the expansion be prevented by increasing the pressure, on the piston, from ‘P’ to ‘P+p’, thus keeping its volume constant. As a result of this, the temperature of gas increases, P-V diagram for an isochoric change is a vertical line parallel to P-axis as shown in below figure.

(e) Quasi-static Process

The process in which change in any of the parameters take place at such a slow speed that the values of P,V, and T can be taken to be, practically, constant, is called a quasi-static process.

(f) Cyclic Process

In a system in which the parameters acquire the original values, the process is called a cyclic process.

(g) Free Expansion

Such an expansion in which no external work is done and the total internal energy of the system remains constant is called free expansion.

Solved Problems

Problem 1:-

In a motorcycle engine, after combustion occurs in the top of the cylinder, the piston is forced down as the mixture of gaseous products undergoes an adiabatic expansion. Find the average power involved in this expansion when the engine is running at 4000 rpm, assuming that the gauge pressure immediately after combustion is 1.50 atm, the initial volume is 50.0 cm³, and the volume of the mixture at the bottom of the stroke is 250 cm³. Assume that the gases are diatomic and that the time involved in the expansion is one-half that of the total cycle.

Concept:-

In an adiabatic process, the final pressure p_f is defined as,

p_f = p_iV_i^γ/V_f^γ

    = p_i [V_i/V_f]^γ

The work done W for adiabatic process is defined as,

W = [p_fV_f - p_iV_i]/(γ-1)

Here, initial pressure is p_i, initial volume is V_i and final volume is V_f and adiabatic constant is γ.

First we have to find out final pressure p_f. To obtain p_f, substitute 16.0 atm for p_i, 50.0 cm³ for V_i, 250 cm³ for V_f and 1.40 for γ in the equation p_f = p_i [V_i/V_f]^γ,

p_f = p_i [V_i/V_f]^γ

    =(16.0 atm) [50/250]^1.40

    = 1.68 atm

To obtain the work done W by the gas during the expansion, substitute 16.0 atm for p_i, 50.0 cm³ for V_i,1.68 atm for p_f, 250 cm³ for V_f and 1.40 for γ in the equation W = [p_fV_f - p_iV_i]/(γ-1), we get,

W = [p_fV_f - p_iV_i]/(γ-1)

    = (16.0 atm) (50.0 cm³) – (1.68 atm) (250 cm³)/(1.40-1)

   = [(16.0 atm×(1.01×10⁵ Pa/1 atm)) (50.0 cm³×(10^-6 m³/1 cm³))] –[ (1.68 atm×(1.01×10⁵ Pa/1 atm)) (250 cm³×(10^-6 m³/1 cm³))]/(1.40-1)

=96.0 Pa.m³

= (96.0 Pa.m³) (1 N/m²/ 1 Pa)

= (96.0 N.m) (1 J/ 1 N.m)

= 96.0 J

Thus the average power P involved in this expansion when the engine is running at 4000 rpm will be,

P = (96.0 J) (8000)/(60 s)

   = 12.8×10³ J/s

   = (12.8×10³ J/s) (1 W/1 J/s)

  = 12.8×10³ W

From the above observation we conclude that, the average power P involved in this expansion when the engine is running at 4000 rpm would be 12.8×10³ W.

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Problem 2:-

A sample of gas expands from 1.0 to 5.0 m³ while its pressure decreases from 15 to 5.0 Pa. How much work is done on the gas if its pressure changes with volume according to each of the three process shown in the pV diagram in below figure.

Concept:-

Work is a path function. So work done on the gas depends upon the path. The area under pv diagram gives the work done on the gas between initial state i and final state f.

Work done W is defined as,

W = -pdV

Here p is the pressure and dV is the change in volume.

(1) In the process 1, the work done W₁ will be zero for vertical path.

So W₁ = 0 J

The work done is present only in horizontal path.

Thus the work done W₂ for horizontal path will be,

W₂= -pdV     (negative sign is due to work is done on the gas)

 = - (15 Pa) (4m³)

= (-60 Pa.m³) (1 J/1 Pa.m³)

= -60 J

Therefore the net work done W on the gas will be,

W = W₁ + W₂

= 0 J+(-60 J)

= -60 J

From the above observation we conclude that, the net work done on the gas would be -60 J.

So W₃ = 0 J

The work done is present only in horizontal path.

Thus the work done W₄ for horizontal path will be,

W₄= -pdV     (negative sign is due to work is done on the gas)

= - (5 Pa) (4m³)

= (-20 Pa.m³) (1 J/1 Pa.m³) = -20 J

Therefore the net work done W on the gas will be,

W = W₃ + W₄

 = 0 J+(-20 J) = -20 J

From the above observation we conclude that, the net work done on the gas would be -20 J.

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Problem 3:-

Calculate the work done by an external agent in compressing 1.12 mol of oxygen from a volume of 22.4 L and 1.32 atm pressure of 15.3 L at the same temperature.

Concept:-

In an isothermal process the work done by an external agent in compressing gas from its initial volume (V_i = V₁  ) to its final volume (V_f = V₂) is,

W = -nRT ln V_f / V_i

    = -nRT ln V₂/ V₁ …… (1)

In accordance to ideal gas equation,

PV = nRT        …… (2)

Here P is the pressure of the gas, V is the volume of the gas, n is the number of moles, r is the gas constant and T is the temperature of the gas.

At constant temperature if the gas compress from its initial pressure P₁, intial volume V₁ to its final pressure P₂, final volume V₂, then

P₁ V₁ = P₂ V₂ = nRT       …… (3)

To find work W in terms of P₁ and V₁, substitute P₁ V₁ for nRT in the equation W = -nRT ln V₂ / V₁ ,

W  = -nRT ln V₂/ V₁

    = -P₁ V₁ ln V₂/ V₁

To obtain work done, substitute 1.32 atm for pressure P₁, 22.4 L for initial volume V₁ and 15.3 L for initial volume V₂ in the equation W = -P₁ V₁ ln V₂/ V₁,

W = -P₁ V₁ ln V₂/ V₁

 = -(1.32 atm) (1.01×10⁵ Pa / 1 atm) (22.4 L) (10^-3 m³ / 1 L)  ln ((15.3 L) (10^-3 m³/1 L) /(22.4 L) (10^-3 m³ / 1 L))

 = (1.14×10³ Pa. m³) ( 1 J/1 Pa. m³)

= 1.14×10³ J

From the above observation we conclude that, the work done by an external agent in compressing the oxygen molecule would be 1.14×10³ J.

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Question:-

Discuss the similarities and especially the distinctions between heat, work, and internal energy.

Answer:-

Heat:- It is the agent which produces in us the sensation of warmth and makes bodies hot. It is a form of energy. It flows from higher to lower temperature. In any exchange of heat, heat lost by the body is equal to the heat gained by the cold body. Substances generally expand when heated but the weight remains the same. A certain amount of heat known as latent heat is required to change the state of a body from solid to liquid or from liquid to gas without any change in temperature.

If the internal energy of the system increases, then the sign of change in internal energy of the system is taken as positive and if the internal energy of the system decreases, then the sign of change in internal energy of the system is taken as negative.