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 v_i to a final volume v_f. Then external work done.

dw = ∫ p dv

PV = μRT => p = μRT/v

Isothermal process.

So T = constant

= μRT [log v_f – log v_i]

= μRT log_e [v_f/v_i]

W = 2.3026 μRT log₁₀ [v_f/v_i]

At isothermal process

PV = constant

P₁V₁ = P₂V₂

= P_i /P_f = V_f /V_i

W = 2.3026 μRT log₁₀ [v_f/v_i]

     = 2.3026 μRT log₁₀ [P_f/P_i]

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

PV^r = constant = K

pᵢ vᵢ^r = p_f v_f^r = k

W = 1/ 1 – r [pᵢ vᵢ - p_f v_f]

P_iv_i – μRT_i

P_fv_f = μRT_f

=> W = μR/ r – 1 [T_i – T_f]

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.C_P = (1 + f/2) R = Q/nΔT
• Work done in a isobaric process.w = p (v_f - v_i) = nR (T_f - T_i) = nRΔT