Pressure ( P) > Volume (V)
T, E, 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.
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,
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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
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:
W = ∫ PdV……(1)
For an adiabatic change,
PV r = G or P = G* V - r Putting it in eq..1
W = ∫ G * V - r dV
= G ∫ V - r dV
= G * [ V (1 - r) / (1 - r) ]
= 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)
M = P₁ V₁ r = P₂ V₂ r...putting in eq (2)
W (adia) = 1/ (r - 1) * [( P₁V₁ * V₁ (r -1 ) - P₂V₂ * V₂ (r -1)]
W(adia) = 1 / (r - 1) * [P₁V₁ - P₂V₂)]......(3)
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 PV ^r = K
P(RT/P) r = G
= P (1 - r) * Tr = G (Constant)
For one mole of gas, PV= RT
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.
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, dE (entropy) =dQ/ dT
dQ will be zero because there is no heat transfer, dQ/ dT = 0
d(E) = 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 that 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.
Now, expansion of ideal gas against a constant external pressure:
1-mole gas (P₁, T₁) = 1 mole of gas (P₂, T₂) where Pext= P₂
dQ = 0,
and dW = - P₂ dV
as P (ext) = P₂ (constant)
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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.
Q1: 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, Tf is very high and 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?
For a gas in cylinder A,
Q (heat) = n x CvdT₁, 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).