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]
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
\(W=\int\limits_{{{V}_{1}}}^{{{V}_{2}}}{P}\,dv\)But for a adiabatic process
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