Question

Method 1 of W By integration, make expressions of work done by gas in (a) Isobaric process (p=constant) (b) Isothermal process (pV=constant) (c) Adiabatic process (PV^gamma= constant)

Answer 1

Answer:

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Answer 2

The expression of work done by gas in Isobaric process, Isothermal process  and Adiabatic process  are given below.

Explanation:

(a) Isobaric process

W=∫Vf - Vi pdV = p ∫Vf - Vi dV (as p= constant)

W = p[V] Vf - Vi = p(Vf − Vi)

W = pΔV

(b) Isothermal process

W = ∫Vf - Vi pdV = ∫Vf- Vi (nRT / V) dV (as p= nRT / V)

W = nRT∫Vf - Vi dV / V (as T=constant)

W = nRTIn(Vf - Vi) = nRTIn(pi - pf) (as piVi = pfVf So, Vf / Vi = pi / pf)

(c) Adiabatic process

pV^γ = constant = k (say) = piVi^γ = pfVf^γ

Further, p = k / V^γ = k / V^−γ

W = ∫Vf - Vi pdV = ∫Vf - Vi kV^−γ dV = [kV^γ+1 / −γ+1] Vf - Vi

= kVf^−γ+1 − kVi^−γ+1 / −γ+1 = pfVf^γ Vf^− γ+1 − piVi^γ Vi ^−γ+1 / 1−γ

= pf Vf − pi Vi / 1 − γ = =nRTf − nRTi / 1 − γ =  n RΔT / 1 − γ