Thermodynamics First Law Assignment Help

Thermodynamics - Thermodynamics First Law

Thermodynamics First Law

Consider an ideal gas in a cylinder fitted with a piston. Assume that the piston is fixed at its position and the walls of the cylinder are kept at a higher temperature than that of the gas. The gas molecules strike the wall and rebounds. The average KE of a wall molecule is higher than average KE of a gas molecule. On collision the gas molecule receives some energy form the wall molecule. This increased Ke is shared by other gas molecules also. In this way total internal energy of the gas increases. 

Now consider that the walls of the cylinder are also at the same temperature as that of the gas. As the as molecules collide with the piston coming toward is, the speed of the molecule increases on collision (assuming the molecules increases as the piston is pushed in. thus we see that emery transfer of together. If ΔQ is the heat supplied and ΔW is the work don3e, then the internal energy of the gas must increase byΔQ - ΔW.

Hence ΔU = ΔQ - ΔW

Δ Q = ΔU + ΔW

Is called the first law of thermodynamics

Work done by a gas = PΔV or W = ∫v1^v2PdV     

The first law denies the possibility of cresting or destroying energy. 

Thermal processes

In general thermal processes may be of three types: (a) eversible, (b) irreversible and (c) cyclic. A reversible process means if a process takes up the path AB then on reversing the conditions it comes back by B.A. a thermal process however cannot be reversible. It could be reversible if the change is extremely small (infinitesimally small)

In irreversible process one will not reach back to A if the process AB has occurred.

Thermal processes may be cyclic or irreversible. Change in internal energy in a cyclic process is zero.

Hence,  Δ Q = ΔW

We can divide these processes as 
(a) Isobaric (b) isochoric (c) Isothermal (d) adiabatic 

In isobaric process pressure remains constant and work done 

W = P Δ V = P(V2 – V1)

∴ dQ = dU + pdV

In isochoric process volume remains constant therefore dV = 0

Hence work done is zero 

∴  Δ Q = ΔU

In isothermal process the temperature remains constant. Melting and boiling are examples. Specific heat in isothermal process is .

W = ∫pdV = nRT ∫v2 v1 dV/V = nRT loge V2 / V1

= 2.303 nRT log V2/V1 = 2.303 nRT log P1/P2

Isothermal elasticity =p (bulk modulus)

In an adiabatic process heat is neither allowed to enter nor allowed to escape the system. Specific heat in an adiabatic process is zero

Since dQ = 0 

∴ dU = -pdV

In an adiabatic process 

pVy = constant 

P1 –yTy = constant 
TVy-1 = constant 

Work done in an adiabatic process, 

W = (p1V1 – p2V2)/ (y – 1) = nR (T1 – T2) /y, where y = Cp/Cv

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