**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

\(w=\int\limits_{{{V}_{i}}}^{{{V}_{f}}}{p}\,dv\)PV = μRT => p = μRT/v

\(w=\int\limits_{{{V}_{i}}}^{{{V}_{f}}}{\frac{\mu RT}{V}}\,dv\)Isothermal process.

So T = constant

\(W=\mu RT\int\limits_{{{V}_{i}}}^{{{V}_{f}}}{\frac{dv}{v}}\) \(=\mu RT\left[ \log _{e}^{v} \right]_{{{V}_{i}}}^{{{V}_{f}}}\)= μRT [log v_{f} – log v_{i}]

= μRT log_{e} [v_{f}/v_{i}]

W = 2.3026 μRT log_{10} [v_{f}/v_{i}]

At isothermal process

PV = constant

P_{1}V_{1} = P_{2}V_{2}

= P_{i }/P_{f} = V_{f }/V_{i}

W = 2.3026 μRT log_{10} [v_{f}/v_{i}]

= 2.3026 μRT log_{10} [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 V_{1} to a final volume V_{2}.

We know that

\(W=\int\limits_{{{V}_{1}}}^{{{V}_{2}}}{P}\,dv\)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_{i}v_{i} – μRT_{i}

P_{f}v_{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

1. For this process Charles law is obeyed.

Hence, v α T => (v₁/v₂) = (T₁/T₂)

2. Specific heat of gas during an isobaric process

C_{P} = (1 + f/2) R = Q/nΔT

3. Work done in a isobaric process

w = p (v_{f} – v_{i}) = nR (T_{f} – T_{i}) = nRΔT