thermodynamics

PV=nRT

∆U=nCv∆T =(f/2)nR∆T =. nR∆Tamma-1. = (PfVf-PiVi)amma-1 ,internal energy dependent on temperature

∆W=Fdx=P∆V=P(V2-V1)

∆Q=∆U+∆W or dQ,=dU+dW ,dQ=nCp∆T(for isobaric process),dU=nCv∆T dW=Pdv

postive values (heat added to system dQ=+ve),(work done by the system expansion dw=+ve), increase of internal energy of the system dU=+ve

∆W=-∆U =nCv(T1-T2)=(f/2)nR(T1-T2) =nR(T1-T2)amma-1 =PiVi-PfVfamma-1 , C=0 ,Beta=gamma*P, slope=gamma*slope of isothermal curve

T= constant dT=0 ∆U=0(it is a slow process) example melting or freezing boiling of water or condensation of steam at 100°C equation of state P1V1=P2V2 ,dQ=dW ,slope of curve -P/V=dP/dV, work done in isothermal process W=2.303nRTlog(V2/V1)= 2.303nRTlog(P1/P2),C=infinity,bulk modulus (beta)=P

v1/v2=T1/T2, ∆w =P∆V=nR∆T, C=(1+(f/2))R =gamma*R/(gamma-1) ,beta=0 dU/dQ=Cv/Cp=1amma ,dW/dQ=R/Cp=(1-(1amma))

∆Q=∆U=nCv∆T , equation of state P1/P2=T1/T2 work done=0 Cv=(f/2)R= R/(gamma-1) ,beta=infinite

equation of state Pv^x =constant,TV^(x-1)=constant , P^(1-x)T^(x)=constant, for x=0 process is isobaric ,x=1 process is isothermal,x=gamma process is adiabatic

∆Q=∆U=0,∆W=0

A. CLAUSIUS STATEMENT
It is impossible for a self acting machine unaided by any external agency to transfer heat from a cold reservoir to a hot reservoir. In other words, heat can't flow by itself from a colder to a hotter body.

B. KELVIN-PLANCK STATEMENT
It is impossible for any heat engine to convert all the heat absorbed from a reservoir completely into useful work. In other words, 100% conversion of heat into work is impossible.

REVERSIBLE
Any process is said to be reversible, if it can be reversed such that both the system and surroundings return to their original states, with no other change anywhere else in the universe.

Necessary conditions for a reversible process:-

1. The process must be quasi static. For this the process must be carried out infinitesimally slowly so that the system remains in thermal and mechanical equilibrium with the surroundings throughout.
2. Dissipative forces such as viscosity, friction, inelasticity, etc should be absent.

IRREVERSIBLE PROCESS
Any process which cannot be replaced in the reverse direction exactly is called irreversible process. Spontaneous process of nature are irreversible.

The device, used to convert heat energy into mechanical energy is called a heat engine.

w=Q1-Q2 ,· Engine derives an amount 'Q_1' of heat from the source.
· A part of this heat is converted into work 'W'.
· Remaining heat 'Q_2' is rejected to the sink. Thus Q_1 = W + Q_2 or the work done by the engine is given by W = Q_1 - Q_2
· The energy Q_2 is unavailable in the universe, which causes increase in entropy of universe

It is an ideal reversible heat engine that operates between two temperatures T_1 (source) and T_2 (sink). It was first conceived by a French engineer. Sadi Carnot in 1824. It operates through a series of two isothermal and two adiabatic processes called Carnot cycle. It is a theoretical heat engine with which the efficiency of practical engines is compared.

1. No heat engine operating between two given temperatures can be more efficient than a Carnot engine operating between the same two temperatures.
2. All reversible heat engines operating between the same two temperature reservoirs have the same efficiency, (efficiency of carnot engine is independent of working substance).

In order to obtain a continuous supply of work, the working substance is subjected to the following cycle of quasi static operations known as Carnot cycle. The working substance of the Carnot engine is assumed to be an ideal gas. isothermal expansion heat absorbed or work done =2.303nRTlog(V2/V1) adiabatic expansion of gas W=nR((T1-T2)amma-1), isothermal compression W=2.303nRTlog(v3/v4) adiabatic compression=nR(T1-T2)/(gamma-1)

eta=w/Q1=Q1-Q2/Q1=T1-T2/T1,eta maximum value is 1 if Q2=0 this is impossible according to second law

The refrigerator is just the reverse of a heat engine.
In refrigerator, the working substance extracts an amount of heat Q_2 from the cold reservoir (sink) at a lower temperature T_2. An amount of external work W is done on the working substance and finally an amount of heat Q_1 is rejected to the hot reservoir at a higher temperature T_1.

beta =Q2/W=Q2/(Q1-Q2). =T2/T1-T2 ,beta=(1-eta)/eta