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Thursday, October 27, 2016

Application of First Law of Thermodynamics on Different Process - II

Isothermal process:-
The process in which temperature remains constant is called isothermal process.

But practically isothermal process is impossible, because for a isothermal process to occur there should be infinite conductive material, which is not possible.

For this purpose only we perform a work slowly so that

External work done by an ideal gas in isothermal process:-

At a constant absolute temperature “T”. μ Moles of an ideal gas is expanded from an initial volume vi to a final volume vf. Then external work done.

dw = ∫ p dv


PV = μRT => p = μRT/v


Isothermal process.

So T = constant


= μRT [log vf – log vi]

= μRT loge [vf/vi]

W = 2.3026 μRT log10 [vf/vi]

At isothermal process

PV = constant

P1V1 = P2V2

= Pi /Pf = Vf /Vi

W = 2.3026 μRT log10 [vf/vi]

     = 2.3026 μRT log10 [Pf/Pi]

Adiabatic process:-
In a adiabatic process heat neither enters the system nor leaves the system.

i.e., Q = 0

ΔV = Q – W

= 0 – W

ΔV = - W

If work is done on the system

=> W = - W

=> ΔV = - (-W) = W

So internal energy increases if work is done by the system.

W = +W

=> ΔV = - (+W) = -W

So internal energy decreases.

Work done in a Adiabatic Expansion:-

Let us consider μ moles of an ideal gas expands adiabatically from an initial volume V1 to a final volume V2.

We know that


But for a adiabatic process

We know that

PVr = constant = K


pᵢ vᵢr = pf vfr = k


W = 1/ 1 – r [pᵢ vᵢ - pf vf]

Pivi – μRTi

Pfvf = μRTf

=> W = μR/ r – 1 [Ti Tf]

Isobaric process:-
An isobaric process is one in which volume and temperature of system may changes but pressure remains constant.

Δp = 0
  • For this process Charles law is obeyed.
    Hence, v α T => (v₁/v₂) = (T₁/T₂)
  • Specific heat of gas during an isobaric process.CP = (1 + f/2) R = Q/nΔT
  • Work done in a isobaric process.w = p (vf - vi) = nR (Tf - Ti) = nRΔT

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