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By learning Physics Topics, we can gain a deeper appreciation for the natural world and our place in it.
What are the Differences Between Isothermal and Adiabatic Process?
Adiabatic process: A process in which no heat is exchanged between a system and its surroundings, is called an adiabatic process. The changes in volume, pressure. temperature and other quantities in an adiabatic process are called adiabatic changes.
Let a gas be enclosed inside a cylinder-piston arrangement. The cylinder and the piston arc made of a non-conducting material. This piston can move without friction along the inner walls of the cylinder. When the gas expands rapidly, work is done by the gas. The internal energy of the gas decreases, as it supplies the energy necessary to do the work. But the internal energy of a gas depends on its temperature.
So the temperature of the gas will also decrease. As the walls of the cylinder are non-conducting and the expansion of gas takes place very fast, no heat can enter from the surroundings to stop this fall in temperature. So there will be no heat exchange during this expansion. This is called an adiabatic expansion. The corresponding fall in temperature is known as adiabatic cooling.
Similarly, if the gas contracts rapidly, work is done on the gas. Hence, heat is generated and temperature of the gas increases. As the gas is compressed very fast, the heat evolved will not be transmitted to the surroundings through the non-conducting cylinder and the gas will remain hot. Such type of compression of the gas is called an adiabatic compression and the corresponding rise in temperature is known as adiabatic heating.
In an adiabatic process, no heat is exchanged. So Q = 0. Then from the first law of thermodynamics,
Q = (Uf – Uj) + W or,
0 = (Uf – Ui) + W
or, W = -(Uf – Ui) = -ΔU
i. e., external work done = decrease in internal energy.
For an adiabatic compression, W is negative. This means that work is done on the system by its surroundings. As a result, the internal energy increases.
Adiabatic conditions:
i) In an adiabatic change, temperature does not remain constant. No heat should be exchanged between a sys tem and its surroundings during this process. So the containers should be made of highly non-conclucting materials. Here, the walls of the containers are known as adiabatic walls. A common example is the walls of a thermosflask.
ii) The process should be very fast; no effective heat exchange can take place in that very short time interval. Because of this reason expansion or compression of the gas should take place very fast. So adiabatic process is a very rapid process and any rapid thermal process is usually regarded as an adiabatic process.
We know that even good conduction is a relatively slow process. If we heat one end of a copper rod, the other end is not heated instantly. The heat takes some time to be conducted to the other end. So a very rapid process is adiabatic even when the container is made of a conducting material. For this reason, a bicycle pump made of brass gets adiabatically heated due to very rapid pumping operations.
Relation between volume, pressure and temperature of an ideal gas in an adiabatic process: in an adiabatic process of an ideal gas, the relation pV = constant (Boyle’s law) is not obeyed, because the temperature is no longer a constant. Instead, the relation between volume and pressure becomes
pVγ = constant ………. (1)
where γ = \(\frac{C_p}{C_v}\) = constant for an ideal gas.
For n mol of an ideal gas, pV = nRT
So, p = \(\frac{n R T}{V}\) and V = \(\frac{n R T}{p}\)
Now, pVγ = constant = A (say)
Then, \(\frac{n R T}{V}\)Vγ = A or, TVγ-1 = \(\frac{A}{n R}\)
i.e., TVγ-1 = constant …….. (2)
Again, p\(\left(\frac{n R T}{p}\right)^\gamma\) = A or, Tγp1-γ = \(\frac{A}{(n R)^\gamma}\)
i.e., Tγp1-γ = constant ……. (3)
Relation (1) to (3) relate V, p and T during an adiabatic process of an ideal gas.
Proof of equation pVγ = constant: In section 1.9.1 it has been shown that the first law of thermodynamics for a hydrostatic system can be written as,
dQ = nCvdT + pdV …….. (4)
Again for n mol of an ideal gas,
pV = nRT
or, pdV + Vdp = nRdT
or, dT = \(\frac{p d V+V d p}{n R}\) …… (5)
For an adiabatic process, dQ = 0
∴ From equations (4) and (5) we get,
0 = nCv\(\left(\frac{p d V+V d p}{n R}\right)\) + pdV
or, 0 = CvVdp + (Cv + R)pdV
or, CvVdp = -CppdV [∵ Cp – CV = R]
or, \(\frac{d p}{p}\) = –\(\frac{C_p}{C_y} \frac{d V}{V}\) or, \(\frac{d p}{p}\) + γ\(\frac{d V}{V}\) = 0 [∵ γ = \(\frac{C_p}{C_v}\)]
∴ \(\int \frac{d p}{p}\) + γ\(\int \frac{d V}{V}\) = constant
or, lnp + γlnV = constant
or, lnp + lnVγ = constant
or, pVγ = constant
Adiabatic process of an ideal gas on a pV-diagram: pV diagrams of adiabatic changes arc called adiabatic curves. For a gas of particular mass, these curves are shown in Fig. where S1, S2 and S3 are the entropies of the gas. (Entropy is discussed in section 1.12.1). The relation pVγ = constant indicates that the adiabatic curves on a pV diagram will be different from the isothermal curves. Calculations show that at every point on a pV diagram the adiabatic curve has a greater slope. This means that the adiabatic curves are steeper than the isothermal curves.
Two adiabatic curves can not intersect because a point of intersection means two entropies of a gas, which is impossible. The slopes of an isothermal and an adiabatic curve on a pV-diagram are called isothermal slope and adiabatic slope, respectively.
Relation between isothermal and adiabatic slopes: For an isothermal change, pV = constant
Differentiating, pdV + Vdp = 0
or, \(\frac{d p}{d V}\) = –\(\frac{p}{V}\) = slope of isothermal curve
For an adiabatic change, pVγ = constant
Differentiating, γpVγ-1 dV + Vγdp = 0
or, \(\frac{d p}{d V}\) = -γ\(\frac{p}{V}\) = slope of adiabatic curve
So, adiabatic slope at a point
= γ × isothermal slope at that point
As γ > 1, the adiabatic slope at every point is greater than the isothermal slope.
Work done by an ideal gas in an adiabatic process: The first law can be written (for all processes of an ideal gas) as
dQ = nCvdT + dW
Now, for an adiabatic process, dQ = 0. Then dW = -nCvdT. So for a change in temperature from Ti to Tf in an adiabatic process of an ideal gas, the total work done is,
In relations (6) and (7), cv and \(\frac{R}{\gamma-1}\) are positive quantities.
As a result,
i) For an adiabatic expansion. W is positive. So Ti – Tf is positive, or Tf < Ti. This means that the temperature decreases. This is adiabatic cooling. For this reason, the air coming out of a bursting bicycle or car tyre appears to be cooler. This process is sometimes utilised to liquefy some gases.
ii) For an adiabatic compression, similar arguments show that the temperature increases. This is adiabatic heating. For this reason, a bicycle or football pump becomes hot during air compression due to pumping. however, adiabatic heating has no practical utilisation.
Comparison between isothermal and adiabatic processes:
Isothermal process | Adiabatic process |
1. Temperature does not change. | 1. Temperature changes. |
2. The system exchanges heat with its surroundings. | 2. No heat exchange takes place between the system and its surroundings. |
3. Ideally, the container walls should be good conductors of heat. | 3. Ideally, the container walls should be poor conductors of heat. |
4. A very slow process. | 4. A very fast process. |
5. Boyle’s law pV = constant is obeyed for an ideal gas, as the temperature remains constant. | 5. Boyle’s law is not obeyed as the temperature changes. The effective relation is pVγ = constant, where γ = \(\frac{C_p}{C_v}\) |
6. The internal energy does not change. | 6. The internal energy of the gas decreases during expansion, but increases during compression. |
Numerical Examples
Example 1.
10 mol of an ideal gas is taken through an isothermal process in which the volume is compressed from 40 L to 30 L. If the temperature and the pressure of the gas are 0°C and 1 atm respectively, find the work done in the process. Given: R = 8.31 J ᐧ mol-1 ᐧ K-1.
Solution:
Work done for n mol of gas, due to change in volume in an isothermal process,
W = nRTln\(\frac{V_f}{V_i}\) = nRT × 2.3026 log10\(\frac{V_f}{V_i}\)
= 10 × 8.31 × 273 × 2.3026 log10\(\frac{30}{40}\)
= 10 × 8.31 × 273 × 2.3026 log10\(\frac{3}{4}\)
= 10 × 8.31 × 273 × 2.3026[0.4771 – 0.6020]
= -10 × 8.31 × 273 × 2.3026 × 0.1251
= -6534.91 J
[the negative sign shows that work is done on the gas]
Example 2.
A gas has an initial volume of 1 L at a pressure of 8 atm. An adiabatic expansion takes the gas to a pressure of 1 atm. What will be its final volume? Given γ = 1.5.
Solution:
\(p_1 V_1^\gamma\) = \(p_2 V_2^\gamma\) or, \(V_2^\gamma\) = \(V_1^\gamma \cdot \frac{p_1}{p_2}\)
Here, p1 = 8 atm, V1 = 1 l, p2 = 1 atm
∴ V2 = \(V_1\left(\frac{p_1}{p_2}\right)^{\frac{1}{\gamma}}\) = 1 × \(\left(\frac{8}{1}\right)^{\frac{2}{3}}\) [γ = 1.5 = \(\frac{3}{2}\);\(\frac{1}{\gamma}\) = \(\frac{2}{3}\)]
= \(\left(2^3\right)^{\frac{2}{3}}\) = 22 = 4 L.
Example 3.
Some amount of air, initially at STP, is a adiabatically compressed to \(\frac{1}{5}\)th its initial volume. Determine the rise in temperature. Given, γ = 1.41.
Solution:
In an adiabatic process, TVγ-1 = constant
or, T1\(V_1^{\gamma-1}\) = T2\(V_2^{\gamma-1}\)
or, T2 = T1\(\left(\frac{V_1}{V_2}\right)^{\gamma-1}\)
Here, T1 = 0°C = 273 K; \(\frac{V_1}{V_2}\) = 5 ; γ – 1 = 1.41 – 1 = 0.41
∴ T2 = 273 × (5)0.41
= 528.13 K = (528.13 – 273)°C = 255.13°C
∴ Rise in temperature = 255.13 – 0 = 255.13°C
Example 4.
Some amount of gas at 27°C is suddenly compressed to 8 times its initial pressure. If γ = 1.5, find out the rise in temperature.
Solution:
As the gas is suddenly compressed, the process is adiabatic.
Example 5.
Find out the work done to expand an ideal gas isothermally to twice its initial volume. [HS ’07]
Solution:
If 1 mol of an ideal gas is at a temperature TK, then pV = RT, p = \(\frac{R T}{V}\)
∴ Work done in isothermal process,
W = \(\int_{V_i}^{V_f} p d V\) = \(\int_{V_i}^{V_f} \frac{R T}{V} d V\) = RT\(\int_{V_i}^{V_f} \frac{d V}{V}\) = RT ln\(\frac{V_f}{V_i}\)
[T = constant in isothermal process]
Here, \(\frac{V_f}{V_i}\) = 2 and R = 8.31 J ᐧ mol-1 ᐧ K-1
∴ W = 8.31 × T × ln2 = 8.31 × 0.693 T
= 5.76 T J.
Example 6.
Find out the work done in adiabatic compression of 1 mol of an ideal gas. The initial pressure and volume of the gas are 105 N ᐧ m-2 and 6 L, respectively; the final volume is 2 L; the molar specific heat of the gas at constant volume is \(\frac{3}{2}\)R.
Solution:
Here, Cv = \(\frac{3}{2}\)R
As Cp – Cv = R, Cp = Cv + R = \(\frac{3}{2}\)R + R = \(\frac{5}{2}\)R;
so, γ = \(\frac{C_p}{C_v}\) = \(\frac{5}{3}\)
For 1 mol of the gas, pV = RT, T = \(\frac{1}{R}\)pV;
the fall in temperature is Ti – Tf = \(\frac{1}{R}\)(piVi – pfVf)
For this adiabatic process, \(p_i V_i^\gamma\) = \(p_f V_f^\gamma\)
Example 7.
The molar specific heat of an ideal gas at constant pressure is Cp = \(\frac{5}{2}\)R. Some amount of this gas in a closed container has a volume 0.0083 m3, a temperature 300 K and a pressure 1.6 × N ᐧ m-2. If 2.49 × 104 J of heat is supplied at a constant volume of the gas, find out the final temperature and pressure. Given, R = 8.3 J ᐧ mol-1 ᐧ K-1.
Solution:
Let n mol of the gas be present in the container.
∴ Final pressure is 3.6 × 106 N ᐧ M-2.
Example 8.
8 g oxygen, 14 g nitrogen and 22 g carbon dioxide are mixed in a container of volume 4 l. Find out the pressure of the gas mixture at 27°C. Given R = 8.315 J ᐧ K-1.
Solution:
If n = number of moles in a gas, then pV = nRT.
So, p = \(\frac{n R T}{V}\) = \(\frac{m}{M} \frac{R T}{V}\), where m = mass of gass and M = molecular weight.
Then the pressures due to oxygen, nitrogen and carbon dioxide gases, respectively, are