Adiabatic Process

The adiabatic process is a thermodynamic process in which the pressure, volume, and temperature of the system change, but there is no exchange of heat between the system and the surroundings.

In a rapid and instantaneous process, such as an adiabatic one, there isn't ample time for heat exchange to occur, thus rendering the process adiabatic.

1.0Definition of Adiabatic Process

An adiabatic process is defined as a thermodynamic process during which there is negligible or no exchange of heat between the system and its surroundings.

This can occur either because the process happens so quickly that there is no time for heat exchange, or because the system is well-insulated, preventing any heat transfer. As a result, the change in internal energy of the system during an adiabatic process is solely due to work done on or by the system.

2.0Equation of Adiabatic Process

The equation of an adiabatic process depends on the context, particularly whether the process is reversible or irreversible. For an ideal gas undergoing an adiabatic process, the equation is given by:

PV^γ = constant

Equation for adiabatic process, PV^γ = constant, TV^{γ – 1} = constant, T^γ p^{1 – γ} = constant

Where :

• P is the pressure of the gas
• V is the volume of the gas
• T is temperature of the gas 
• γ is the adiabatic constant or ratio of specific heats

• C_p is the specific heat at constant pressure 

• C_v is the specific heat at constant volume.

For an ideal monatomic gas, $\gamma =\frac{5}{3}$

For an ideal diatomic gas, $\gamma =\frac{7}{5}$

This equation reflects the conservation of energy in an adiabatic process, where no heat is exchanged with the surroundings, so the change in internal energy (ΔU) is entirely due to work done on or by the gas.

3.0Work Done in Adiabatic Process

Let initial state of system is (P₁, V₁, T₁) and after adiabatic change final state of system is (P₂, V₂, T₂) then we can write $P_1{V}_1=P_2{V}_2=K$ (here K is constant)

$\mathrm{W}={\int }_{V_1}^{V_2}PdV=K{\int }_{V_1}^{V_2}{V}^{-\gamma }dV=K{\left(\frac{V^{-\gamma +1}}{-\gamma +1}\right)}_{V_1}^{V_2}=\frac{K}{(-\gamma +1)}\left[V_2^{-\gamma +1}-V_1^{-\gamma +1}\right]\left(\because K=P_1{V}_1^{\gamma }=P_2{V}_2^{\gamma }\right)$

So

 $\Rightarrow W=\frac{1}{(\gamma -1)}\left[P_1{V}_1^{\gamma }{}^{\gamma }{V}_1^{-\gamma }\cdot V_1-P_2{V}_2^{\gamma }{V}_2^{-\gamma }\cdot V_2\right]=\frac{1}{(\gamma -1)}\left[P_1{V}_1-P_2{V}_2\right]\Rightarrow W=\frac{\mu R}{(\gamma -1)}\left(T_1-T_2\right)(\because PV=\mu RT)$

Form of first law : dU = – δW 

It means the work done by an ideal gas during adiabatic expansion (or compression) is on the cost of change in internal energy proportional to the fall (or rise) in the temperature of the gas.

If the gas expands adiabatically, work is done by the gas. So, W_adia is positive.

The gas cools during adiabatic expansion and T₁ > T₂.

If the gas is compressed adiabatically, work is done on the gas. So, W_adia is negative.

The gas heats up during adiabatic compression and T₁ < T₂.

4.0Graphs in Adiabatic Process

Slope of the adiabatic curve

For an adiabatic process, PV^γ = constant               

Differentiating, γPV^{γ – 1} dV + V^γ dP = 0

⇒ V^γ dP = –γPV^{γ – 1} dV

⇒ $\frac{\mathrm{dP}}{\mathrm{dV}}=-\frac{\gamma {\mathrm{PV}}^{\gamma -1}}{{\mathrm{V}}^{\gamma }}=-\gamma \frac{\mathrm{P}}{\mathrm{V}}=\gamma \left(-\frac{\mathrm{P}}{\mathrm{V}}\right)$

Slope of adiabatic curve,

${\left[\frac{\mathrm{dP}}{\mathrm{dV}}\right]}_{\text{adiabatic }}=-\frac{\gamma \mathrm{P}}{\mathrm{V}}$

The magnitude of the slope of the adiabatic curve is greater than the slope of the isotherm

${\left|\frac{\mathrm{dP}}{\mathrm{dV}}\right|}_{\text{adiabatic }}=\gamma \frac{\mathrm{P}}{\mathrm{V}}=\gamma {\left|\frac{\mathrm{dP}}{\mathrm{dV}}\right|}_{\text{isothermal }}\Rightarrow \frac{\text{ slope of adiabatic changes }}{\text{ slope of isothermal changes }}=\gamma$

Since γ is always greater than one so an adiabatic curve is steeper than an isothermal curve.

5.0Solved Examples of Adiabatic Process

• A gas enclosed in a thermally insulated cylinder fitted with a non–conducting piston. If the gas is compressed suddenly by moving the piston downwards, some heat is produced. This heat cannot escape from the cylinder. Consequently, there will be an increase in the temperature of the gas.
• If a gas is suddenly expanded by moving the piston outwards, there will be a decrease in the temperature of the gas.
• Bursting of a cycle tube.
• Propagation of sound waves in a gaseous medium.
• In diesel engines, burning of diesel without a spark plug is done due to adiabatic compression of diesel vapour and air mixture.

Example:

The temperature and pressure of air in the tyre tube are 27°C and 8 atm, respectively. If the tube suddenly bursts, then calculate the final temperature of the air. [γ = 1.5)

Solution:

⇒ ${\mathrm{P}}_1^{1-\gamma }{\mathrm{T}}_1^{\gamma }={\mathrm{P}}_2^{1-\gamma }{\mathrm{T}}_2^{\gamma }\Rightarrow (8{)}^{1-\gamma }(300{)}^{\gamma }=(1{)}^{1-\gamma }\left(T_2\right)\gamma$

⇒ T₂ = 150 K ⇒ T₂ = 150 – 273 = – 123°C

Example:

A diatomic ideal gas is compressed adiabatically to $\frac{1}{32}$ of its initial volume. If the initial temperature of the gas is T₁K and final temperature is aT₁ then find the value of a.

Solution:

${\mathrm{T}}_1{\mathrm{V}}_1^{\gamma -1}={\mathrm{T}}_2{\mathrm{V}}_2^{\gamma -1}\Rightarrow {\mathrm{T}}_1(\mathrm{V}{)}^{\frac{7}{5}-1}=\mathrm{a}\cdot {\mathrm{T}}_1{\left(\frac{\mathrm{V}}{32}\right)}^{\frac{7}{5}-1}\Rightarrow \mathrm{a}=2^{5\times \frac{2}{5}}=4$

Example:

An ideal gas $\left(\gamma =\frac{5}{3}\right)$is adiabatically compressed from to 80 cm³ to 10 cm³. If the initial pressure is P then find the final pressure?

Solution:

$\therefore {\mathrm{P}}_1{\mathrm{V}}_1^{\gamma }={\mathrm{P}}_2{\mathrm{V}}_2^{\gamma }\Rightarrow \frac{{\mathrm{P}}_1}{{\mathrm{P}}_2}={\left(\frac{{\mathrm{V}}_2}{{\mathrm{V}}_1}\right)}^{\gamma }\Rightarrow \frac{\mathrm{P}}{{\mathrm{P}}_2}={\left(\frac{10}{80}\right)}^{\frac{5}{3}}\Rightarrow {\mathrm{P}}_2=32\mathrm{P}$