Thermodynamics is the branch of science that deals with the different forms of energy, the quantitative relationships between them and the energy changes that occur in the physical and chemical process.

Limitations of Thermodynamics:

  • Thermodynamics deal with macroscopic properties like pressure, temperature, volume, and composition. It does not concern with atomic structure and molecular structure.
  • It can predict whether the chemical reaction can occur under given set of conditions but it is not able to explain the rate of reaction.
  • It is more concerned with the initial and final states of the system and not the mechanism of the process.

Terms used in Thermodynamics:

System:

  • The portion of the physical universe under thermodynamic study is called the system.

Surroundings:

  • Remaining part of the universe under thermodynamic study is called the surroundings.

Boundary:

  • The real or imaginary surface separating the system from the surrounding is called the boundary.

Chemical Thermodynamics 01

Different Types of Systems On the Basis of Exchange of Matter and Energy: 

Open System :

  • A system which can exchange both energy and matter with the surroundings is called an open system.
  • Example: A hot solution in a beaker can exchange both the matter  (vapours) and energy (heat) to the surroundings.

Closed system:

  • A system which can exchange energy but not matter with the surroundings is called a closed system.
  • Example: Water or gas in a closed (sealed) vessel. The substance can be heated or it can give out heat (energy exchange) but no substance can escape from the vessel.

Isolated system:

  • A system which can exchange neither matter nor energy with the surroundings is called isolated system.
  • Example: A liquid placed in a thermos flask is an isolated system. Temperature change outside the flask does not change the temperature of the liquid. (no energy transfer) and nothing can escape from or enter the flask (no transfer of matter).

Thermodynamics 002

Different Types of Systems On the Basis of Phases of Matter (composition):

Homogeneous system:

  • When a system is uniform, throughout or consists of a single phase, it is said to be the homogeneous system.
  • Examples: A pure single solid, liquid or a gas. A mixture of gases. The true solution of solid in liquid.

Heterogeneous system:

  • A heterogeneous system is one which is not uniform throughout and which contains two or more phases which are separated from one another by definite boundary surface.
  • Examples: Mixture of two immiscible liquids such as benzene and water, The mixture of two or more solids.

Properties of System:

Extensive Properties:

  • Extensive property of a system is defined as a property that depends on the amount (or amounts) of the substance (or substances) present in the system.
  • Examples: Mass, Volume, and Energy are extensive properties.

Intensive Properties:

  • An intensive property of a system is defined as the property which is independent of the amount of the substance.
  • Examples: Refractive index, density, surface tension.

State of a System:

  • To describe the system completely and ambiguously, macroscopic properties such as pressure, volume, temperature, mass (number of moles) and composition are used. By assigning numerical values to these properties, the state of a system can be defined.

State Function:

  • Any property of a system whose value depends on the current state of the system and is independent of the path followed to reach that state is called the state function.
  • Examples: Pressure (P), volume (V), Temperature  (T) Internal energy (E) Enthalpy (H).

Path Function:

  • The property of a system whose value depends on the path used to reach particular state is called path function.
  • Examples: Work (W), Heat (q)

Thermodynamic Equilibrium:

  • A system is said to be in a state of thermodynamic equilibrium when the state functions of the system do not change with time.

Different Types of Thermodynamic Equilibrium:

  • Thermal Equilibrium:
    • A system is said to be in thermal equilibrium with the surroundings when system and surroundings are at the same temperature and there is no exchange of heat energy between them.
  • Chemical Equilibrium:
    • A system is said to be in chemical equilibrium when the chemical composition of reactants and products do not change with time. Thus the chemical composition of a system as a whole remains constant
  • Mechanical Equilibrium:
    • A system is said to be in mechanical equilibrium when net force acting on the system is zero and net moment of the system is zero.

Pressure – Volume Work:

Concept of Work 01

  • Consider an ideal gas having definite mass (say n moles) be enclosed in a cylinder fitted with weightless, frictionless, tightly fitted, movable piston. Let ‘A’ be the area of the cross-section of the cylinder. Let the gas expand from volume V1 to V2 against constant external pressure Pext which is exerted on the piston. Due to expansion, the piston moves upward by a distance ‘d ‘ .

Thermodynamics 003

Free Expansion:

  • The free expansion is also known as work of expansion in a vacuum. When a gas expands in a vacuum there is no opposing force i.e.  Pext  = 0. Hence work done in free expansion is zero.

Cyclic Process:

  • A cyclic process is one which consists of a series of intermediate steps, at the end of which the system returns to its initial state.
  • Since in a cycle, initial and final states are the same. ΔU  =  0. Hence q  =  -W

Problems involving Pressure Volume Work

  • 2 moles of an ideal gas are expanded isothermally from volume of 15.5 L to the volume 20 L against a constant external pressure of 1 atm. Calculate the pressure volume work in L atm and J (Ans:  – 4.5 L atm or – 455.8 J)
  • 2 moles of an ideal gas are compressed isothermally from volume of 10 dm3 to the volume 2 dm3 against a constant external pressure of 1.01 × 105 Nm-2,. Calculate the pressure volume work done.(Ans: +  808 J)
  • 3 moles of an ideal gas are expanded isothermally from volume of 300 cm3 to the volume 2.5 L against a constant external pressure of 1.9 atm at 300 K. Calculate the pressure volume work in L atm and J.(Ans: – 4.5 L atm or – 423.4 J)
  • 1 mole of an ideal gas is compressed isothermally from volume of 500 cm3 against a constant external pressure of 1.216 × 105 Pa. The pressure volume work involved in the process is 36.5 J. calculate the final volume. (Ans: 200 cm3)
  • 1 mole of an ideal gas is compressed isothermally from volume of 20 L to 8 L against constant externa pressure, when pressure volume work obtained is 44.9 L atm. Find the constant external pressure. (Ans: 3.74 atm)
  • One mole of a gas expands by 3 L against a constant pressure of 3 atmosphere. Calculate the pressure volume work done in a) L-atm b) joules and c) calories. (Ans: – 9  L atm or – 911.7 J  or -217.9 cal)
  • 100 mL of ethylene(g) and 100 mL of HCl (g) are allowed to react at  2 atm pressure as per the reaction given below. Calculate pressure volume type of work in it in joules.

C2H4(g)    +  HCl(g)    →     C2H5Cl(g)                       (Ans: + 20.26 J)

  • A gas cylinder of 5 L capacity containing 4 kg of helium gas at 27 °C developed a leakage leading to the escape of the gas into atmosphere. If atmospheric pressure is 1.0 atm. calculate the pressure volume work done by the gas assuming ideal behaviour. (Ans: – 2494 kJ)
  • 1.6 mol of water evaporates at 373 K against atmospheric pressure of 1 atm. Assuming ideal behaviour of water vapours calculate the work done. (Ans:  – 4954.8 J)
  • 1.0 mol of water evaporates at 373 K against atmospheric pressure of  749.8 mm of Hg. Assuming ideal behaviour of water vapours calcul;ate the pressure volume work done. (Ans: – 3101 J)

For Solution of the Problems Click Here

Concept of Maximum Work:

  • According to the first law of thermodynamics, ΔU = q  +  W    In an isothermal process, ΔU  =   0, ∴   q   =   – WTherefore, all the heat absorbed by the system is utilized to do work.
  • Work of expansion is given by the product of external pressure and the volume change. W   =   – Pext ΔV
  • In any expansion, the external pressure must be less than the pressure of the gas. If the external pressure is zero, the work done is also zero as the gas expands into the vacuum. If the external pressure is increased gradually, more and more work will be done by the gas during expansion. When Pextbecomes P then ΔV will be maximum. In a eversible process, Pext is always less than the pressure of the gas, by an infinitesimally small quantity.

W  =  (P  –  dp ) dV

  • To get maximum work the quantity dP should be very very small. Therefore work done in an isothermal reversible expansion of an ideal gas is maximum work.

Expression for Maximum Work:

  • Work done in an isothermal reversible expansion is maximum work.

Concept of Work 01

  • Consider  ‘n ‘ moles of an ideal gas enclosed in a cylinder fitted with a weightless, frictionless, air tight movable piston. Let the pressure of the gas be P which is equal to external atmospheric pressure P. Let the external pressure be reduced by an infinitesimally small amount dp and the corresponding small increase in volume be dV. so the small work done in the expansion process

Thermodynamics

  • When the expansion of the gas is carried out reversibly then there will be series of such p.dV terms. The total maximum work  Wmax can be obtained by integrating this equation between the limits V1 to  V2. Where V1 is initial volume and  V2 is final volume.

Work Done 04

For an isothermal expansion , Boyle’s law is applicable.
Hence P1V1 = P2V2 i.e. V2 / V1 = P1 / P2

Work Done 05

Where  P1  and P2  are initial and final pressure respectively.

Work is Path Function:

  • The work done in isothermal constant pressure process is given by W   =  – Pext  ( V2  –  V1)
  • The work done in the isothermal reversible process is given by Thermodynamics 001
  • Thus the work done depends on the manner or the conditions or the path under which the process carried out. Hence work is path function or it is not a state function.

Sign Convention Used in Thermodynamics:

  • Work of expansion is taken as negative (- W).
  •  Work of compression is taken as positive (+W).

Problems Based on Isothermal Reversible Process and Maximum Work

  • 3 moles of an ideal gas are expanded isothermally and reversibly from volume of 10 m3 to the volume 20 m3 at 300 K. Calculate the work done. (Ans:  – 5.187 kJ)
  • 24 g of oxygen are expanded isothermally and reversibly from 1.6 × 105 Pa pressure to 100 kPa  at 298 K. Calculate the work done. (Ans: – 873.4 J)
  • 4.4 × 10-2 kg of CO2 is compressed isothermally and reversibly at 293 K from the initial pressure of 150 kPa when work obtained is 1.245 kJ. Find the final pressure.(Ans:  250 kPa)
  • 2.8 × 10-2 kg of nitrogen is expanded isothermally and reversibly at 300 K from the initial pressure of 15.15 × 105 Nm-2 when work obtained is 17.33 kJ. Find the final pressure. (Ans: 14056.7 Nm-2)
  • 3 moles of an ideal gas are compressed isothermally and reversibly at 22° C to a volume 2L when work done is 2.983 kJ. Find the initial volume. (Ans:  4 L)
  • 280 mmol of an ideal gas occupy 12.7 L at 310 K. Calculate the work done when gas expands a) isothermally against constant external pressure 0.25 atm, b) isothermally and reversibly c) in a vacuum until its volume becomes 3.3 L. (Ans: -38.6 J, -166.9 J, 0 J)
  • 300 mmol of a perfect gas occupies 13 L at 320 K. Calculate the work done in joules when the gas expands a) isothermally against the constant external pressure of 0.20 atm. b) Isothermal and reversible process c) Into vacuum until the volume of the gas is increased by 3 L. (Ans: – 60.78 J, -165.8 J, 0 J)

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