Introduction
A thermodynamic system in which there is a change in the state of matter due to the change in the Pressure, Volume, Temperature (P, V, T) without transferring heat or mass with the thermodynamic system or its surroundings.
The process in which the work done is in the form of a change in internal energy (U) and the amount of heat transferred, Q is zero (or there is no gain or loss of heat)
On this page, we shall learn about the various topics:
Adiabatic process
Related examples
Derivation of the Adiabatic Process formula
Work done
Adiabatic expansion and compression
Adiabatic- Reversible and Irreversible Process.
There are four types of processes in a thermodynamic system, which are shown via an image below:
(image will be uploaded soon)
Adiabatic Process: It can be considered as an ideal process in which there is no heat transfer and the change in internal energy would be equal to the work done.
Here, ∆U = U₂ - U₁ ,and W =- ∆U = U1 - U2
W = - ∆U
Since there is no heat transfer that's why the Entropy (∆S) = 0.
Used in -
Turbines, Hot water flask, compression, and expansion of charge in IC.
Adiabatic Compression and Expansion
Adiabatic Compression
Pressure ( P) > Volume (V)
T, S, and U increase during compression.
In cylinders of a car, the compression of the gas-air mixture takes place in no time that there could be an exchange of heat between them.
Adiabatic Expansion
The work is done by the gas in expanding the volume and there's a decrease in the temperature.
Let's take an example here,
(image will be uploaded soon)
In picture 1, the molecules of a gas are tightly bound together as soon as the membrane is punctured, the volume of the gas expands to Volume dV, and a drop in temperature by dT.
In this case, the work done by a gas W = PdV and dQ = 0.
For an Adiabatic Process for an ideal gas is given by:
PV r = Constant
Work Done in an Adiabatic Expansion
Consider a mole of gas contained in a cylinder having insulating walls, provided with a frictionless and insulating piston
Let P be the pressure of the gas when the piston moves up through a small distance dx, the work done will be:
dW = PAdx = PdV
Where A is the cross-sectional area of the piston and dV = Adx is the increase in the volume.
As the gas expands adiabatically, there will be a change in its initial stage (P1, V1, T1) to (P2, V2, T2).
The total work done will be:
v₂
W = ∫ PdV……(1)
v₁
For an adiabatic change,
PV r = G or P = G* V - r Putting it in eq..1
$w = \int_{v1}^{v2}G*V^{-r} dv$
$= G \int_{v1}^{v2}V^{-r}dv$
v₂
= G * [ V(1-r)1-r ]
v₁
= G/ (1-r) * [ V₂ (1 - r) - v₁ (1 - r)]
= G/ (r -1) * [v₁ (1 - r) - v₂ ( 1 - r)]
= 1/ (r -1) * [G* v₁ (1 - r) - G * v₂ (1 - r)] …….(2)
G = P₁ V₁ r = P₂ V₂ r...putting in eq (2)
W (adia) = 1/ (r - 1) * [( P₁Vr₁ * V₁ (1-r ) - P₂Vr₂ * V₂ (1-r)]
W(adia) = 1 / (r - 1) *
(P1V1−P2V2)
P1V1−P2V2)......(3)
Deriving an Adiabatic Equation
Adiabatic relationship between P and V
According to the first law of thermodynamics:
dQ = dU + dW….(1)
For one mole of gas, the equation is:
dW = PdV
dU = nCVdT, and CV = dU/dT => dU = nCVdT = CVdT (as n=1)
Also, for an adiabatic process, dQ = 0;
According to the ideal gas equation:
PV= RT.....(2) (n =1)
Now putting the values of dU and dW in equation(1):
0 = CVdT + PdV…..(3)
Differentiating both the sides in equation (2), we get:
PdV + VdP = RdT
dT= (PdV + VdP)/ R
Putting the value of dT in q..(3)
CV (PdV + VdP)/ R + PdV = 0
CV (VdP) + CV + R (PdV)…(4)
As we know that CP = CV + R putting it in eq..(4)
CV (VdP) + CP (PdV) = 0
Dividing both the sides by CVPV, we get:
dP/P + (Cp/Cv)* dV/V = 0
We know that Cp / Cv = r
dP/P + r* dV/V = 0…….(5)
Integrating both the sides in eq…(5)
= ∫ dP/P + r ∫ dV/V = C (∵ C, an integral constant)
= Loge P + r Logₑ V = C
= Loge PV r = C
= PV r = ec
= PV r = G
Adiabatic Relation Between P and T
We already know that for an ideal gas, the equation is given by,
PV = RT ( for one mol gas)
V = RT/P…(1)
Putting the value of equation (1)....in the equation PVr= K
P(RT/P) r = G
= P (1 - r) Tr = G (Constant)
Adiabatic Relation Between V and T
For one mole of gas, PV= RT
P =RT/V
Putting PV r =G, we get
RT/V * Vr = G or T*V(r - 1) = G/ R
= TV(r - 1) = G (Constant)
This equation describes the adiabatic relation between V and T for an ideal gas.
Adiabatic - Reversible and Irreversible process
Reversible Adiabatic Process
A reversible adiabatic process is basically an isentropic process.
What is an Isentropic Process?
Since in an adiabatic process, the heat transfer is zero which means the change in entropy is also zero. That’s why the process is said to be isentropic by nature.
As, S (entropy) =dQ/ dT
dQ will be zero because there is no heat transfer, dQ/ dT = 0
d(S) = 0
Let’s Consider Some Real-Life Examples:
Expansion of steam in steam turbines.
Gas in gas turbines
Compression of air in compressors
PVr is constant along a reversible adiabatic process.
Irreversible Adiabatic Process
As the name suggests, the process can’t be traced back to its original state.
During an irreversible adiabatic process of expansion. There will be a change in entropy because of frictional dissipation.
Irreversible expansion cannot be performed at equilibrium.
(image will be uploaded soon)
Summary
An Irreversible process can't be in a quasi-static state (the process that happens slowly to keep it in an equilibrium)
A reversible process is a quasi-static process.
Application-Based Questions
1. Let's take an example of the daily based activity, of pumping air inside the tire of a bicycle using a hand pump. Here, we would consider air inside the pump is a thermodynamic system having Volume V at an atmospheric pressure P, and the room temperature, 30°C. Suppose the nozzle of the tire is blocked and you push the pump to ⅕ of the volume V. Calculate the final temperature of the air in the pump.
Solution: Since the nozzle of the tire is blocked air will not flow, we will consider this as an Adiabatic Process
Ti Vi (r -1) = Tf Vf (r - 1)
Ti = 30°C = (273 + 30) K = 303 K
Vi = V , so Vf = V/5
Tf = Ti *(Vi / Vf)(r -1)
= 303 K* 5(1.4 -1)
= 303 K * 50.4
= 303 * 1.90365
Tf = 576.81 K
Since the temperature T is very high it is dangerous to touch the nozzle.
Q2: Two cylinders A and B of equal size (both fitted with piston) are filled with an equal amount of ideal gas at room temperature. In-cylinder A the piston is free to move, while in B, it is fixed. When some amount of heat is added to cylinder A raises by 30 K. What will be the rising temperature of the gas in cylinder B?
Solution: For gas in cylinder A,
Q (heat) = n x CpdT1, and that for cylinder B = n x CvdT2
So, dT2 = Cp/ Cv x dT1
= 7/5 x 30.
(Here, Cp/ Cv = 7/5)
dT2 = 42 K (dT2 = Rise in temperature in cylinder B).
FAQs on Adiabatic Process
1. What is an adiabatic process in thermodynamics?
An adiabatic process is a thermodynamic process where there is no heat or mass transfer between a system and its surroundings. Essentially, the system is perfectly insulated. Any work done on or by the system directly results in a change to its internal energy, leading to a change in temperature, pressure, or volume.
2. What is the main difference between an adiabatic and an isothermal process?
The primary difference lies in what remains constant during the process. In an adiabatic process, the heat exchange (Q) with the surroundings is zero, but the temperature can change. In an isothermal process, the temperature (T) of the system is held constant, which requires heat to be transferred in or out of the system to maintain it.
3. What is the fundamental equation for an adiabatic process in an ideal gas?
For an ideal gas undergoing a reversible adiabatic process, the key relationship between pressure (P) and volume (V) is given by the equation: PVγ = Constant. Here, γ (gamma) represents the adiabatic index or the ratio of specific heats (Cp/Cv), which is a constant value for a specific gas.
4. Can you provide a real-world example of an adiabatic process?
A common example is the rapid compression of air in a bicycle pump. When you push the pump handle down quickly, the air is compressed so fast that it has little to no time to exchange heat with the pump's walls. This adiabatic compression does work on the air, increasing its internal energy and temperature, which is why the pump's nozzle gets hot.
5. If no heat is added, how does the temperature of a gas increase during adiabatic compression?
The temperature increases because work is done on the gas. According to the First Law of Thermodynamics (ΔU = Q - W), since heat transfer (Q) is zero in an adiabatic process, the change in internal energy (ΔU) equals the work done on the gas. During compression, this external work directly increases the gas's internal energy, which manifests as a rise in its kinetic energy and, therefore, its temperature.
6. What happens to the internal energy and temperature during adiabatic expansion?
During an adiabatic expansion, the gas performs work on its surroundings (for example, by pushing a piston). As the system is insulated, it cannot draw heat from the outside to perform this work. Instead, it uses its own internal energy. This expenditure of internal energy results in a decrease in both the temperature and internal energy of the gas, causing it to cool down.
7. Why is a reversible adiabatic process also known as an isentropic process?
A process is called isentropic if the entropy of the system remains constant. A reversible adiabatic process fits this description perfectly because:
- Adiabatic: It involves no heat transfer (Q = 0). Since change in entropy is related to heat transfer (ΔS ∝ Q), this means no entropy flows in or out of the system.
- Reversible: The process is ideal and frictionless, meaning no new entropy is generated internally.
8. How is the work done calculated in a reversible adiabatic process?
The work done (W) by a gas as it changes its state from an initial (P₁, V₁) to a final (P₂, V₂) state in a reversible adiabatic process is calculated by the formula: W = (P₁V₁ - P₂V₂)/(γ - 1). Using the ideal gas law (PV = nRT), this can also be expressed in terms of the initial and final temperatures (T₁ and T₂) as: W = nR(T₁ - T₂)/(γ - 1).
