# 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 thatExternal 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

$$w=\int\limits_{{{V}_{i}}}^{{{V}_{f}}}{p}\,dv$$

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 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]

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

$$W=\int\limits_{{{V}_{1}}}^{{{V}_{2}}}{P}\,dv$$

We know that

PVr = constant = K

$$W=\int\limits_{{{V}_{i}}}^{{{V}_{f}}}{p}\,dv=\int\limits_{{{V}_{i}}}^{{{V}_{f}}}{\frac{k}{{{v}^{r}}}}\,dv\,\,\,\,\left[ p=\frac{k}{{{v}^{r}}} \right]$$ $$k\int\limits_{{{V}_{i}}}^{{{V}_{f}}}{-{{v}^{r}}}\,dv=\frac{k}{1-r}\left[ {{v}^{1-r}} \right]_{{{v}_{i}}}^{{{v}_{f}}}$$ $$=\frac{1}{1-r}\left[ v_{f}^{1-r}-v_{i}^{1-r} \right]$$ $$=\frac{1}{1-r}\left[ \frac{k}{v_{i}^{r-1}}-\frac{k}{v_{f}^{r-1}} \right]$$

pᵢ vᵢr = pf vfr = k

$$W=\frac{1}{1-r}\left[ \frac{{{p}_{i}}v_{i}^{r}}{v_{i}^{r-1}}-\frac{{{p}_{f}}v_{f}^{r}}{v_{f}^{r-1}} \right]$$

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

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

CP = (1 + f/2) R = Q/nΔT

3. Work done in a isobaric process

w = p (vf – vi) = nR (Tf – Ti) = nRΔT