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**Work:**

Imagine the hydrostatic system contained in the cylinder with movable piston. Assume external force F acted in direction showed moving piston from initial point 1 to final point 2 through distance dx. Assume that cylinder has cross section area A, that pressure applied on system at piston face is P, and that force is PA. System also applies opposing force on piston. Work done dW on system in the method described above is

dW = PAdx

But Adx = dV

Therefore dW = -PdV

Negative sign in the last equation specify negative change in volume (that is decrease in volume).

In the finite quasi-static procedure in which volume changes from V_{i} to V_{f}, work done is

W = -∫_{vi}^{vf}PdV

**Work in Quasi-Static Process:**

For the quasi-static isothermal expansion of compression of the ideal gas

W = -∫_{vi}^{vf}PdV

But the ideal gas is one which equation of state is PV = nRT , where n and R are constant. Replace P with nRt/V in equation

W = -∫_{vi}^{vf}(nRt/V)dV

And as T is constant for isothermal process,

W = -nRt∫_{vi}^{vf}VdV

Integration provides

W = -nRtlnV_{f}/V_{i}

**Work and Internal Energy:**

When the adiabatic work (dW_{ad}) is done on or by system, the internal energy of the system changes. Change in internal energy (ΔU ) is equal to adiabatic work done.

dU = dw_{ad}

If system changes from state 1 to state 2 by doing adiabatic work, and if states are varied by finite amount, then

∫_{U1}^{U2}dU = U_{2} - U_{1} = ∫_{1}^{2}dW_{ad} = -W_{ad}

Assume work done is mechanical work (i.e. mechanical adiabatic work), then

U_{2} - U_{1} = -∫_{1}^{2}PdV

*Heat:*

Heat (Q) is a form of energy that is transferred from one part of system to another or to another system by virtue of difference in temperature. Temperature gradient determines direction of heat flow. When heat flows in or out of system from its surroundings, temperature of the system increases or decreases. And internal energy of system changes from initial state (U_{i} ) to final state (U_{f} ). This change in internal energy DU should be equal to the heat flow i.e.

ΔU = U_{f} - U_{i} = Q

Sign of Q

Q is positive when there is the flow of heat in system

Q is negative when there is flow of heat from system

Heat bath or heat reservoir: This is a body which is so large that its temperature doesn't change significantly when heat flows in or out of it.

Heat sink: Just like heat bath, this is a body which is so large that its temperature doesn't change significantly when heat flows in or out of it. Temperature of heat sink is lower compare to that of heat bath.

*First Law of Thermodynamics:*

The internal energy of the system can change only if:

a) There is flow of heat in system or out of the system

b) Work is done on system or by system.

Therefore, change in internal ΔU of system is

ΔU = U_{f} - U_{i} = Q - W

Where Q is heat and W is work. This equation is first law of thermodynamic. Differential for of first law of thermodynamics is

DU = Dq - PdV

Statement of First Law of Thermodynamics

The internal energy of the system tends to increase if energy is added as heat Q and tends to decrease if energy is lost as work W done by system.

*Response Functions:*

When heat is added or withdrawn from the system, there is change in one, two or all its properties. This change in measurable property/properties (macroscopic behavior) is/are basis of thermometer. We can characterize macroscopic behavior of the system response's functions. These functions can be estimated experimentally from changes in thermodynamic coordinates by use of external probes. Examples of response functions are Heat Capacities, force constant (like isothermal compressibility), and thermal response (like expansivity of a gas).

**Heat Capacities:**

When heat is added to the system, its temperature will change. Heat capacities are attained from change in temperature of the system on addition of heat to system.

From equation of first law, dQ = dU + PdV

Heat capacity at constant volume

C_{v} = (∂Q/∂T)v

Heat capacity at constant pressure

C_{p} = (∂Q/∂T)P

Heat capacities can be estimated experimentally.

**Force Constant:**

Force constants estimate (infinitesimal) ratio of displacement to force and are generalization of spring constant. Examples comprise isothermal compressibility of a gas

kT = -1/V [(∂V/∂P)T]

And susceptibility of the magnet

χT = 1/V[(∂M/∂B)T]

It can be shown using equation of state for ideal gas ( PV ∝ T ) that

kT = 1/P

**Thermal Response:**

This probes change in thermodynamic coordinate with temperature. Example, coefficient of volume expansion (expansivity of a gas) is given by

β = 1/V[(∂V/∂T)P]

And this is equal to 1/T for ideal gas.

*Special Cases of First Law of Thermodynamics:*

Consider four different thermodynamic procedures in which certain limitation is imposed on system and corresponding implication when apply to first law. Adiabatic process, dQ = 0, and equation of first law reduces to

dU = dW = PdV

Isochoric process, dV = 0, and equation of first law becomes

dU = dQ = cVdT

Cyclic Process, dU = 0, and equation of first law becomes

Q = W

Free expansion, Q = W = 0, and equation of first law becomes dU = 0

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## Normal Forms

Theory and lecture notes of Normal Forms all along with the key concepts of normal forms, Chomsky normal form, reibach normal form, Context Free Grammars & Languages. Tutorsglobe offers homework help, assignment help and tutor’s assistance on Normal Forms.