Adiabatic process

Adiabatic Compression

In Greek, adiabatic means “nothing passes through”. The process in which pressure, volume and temperature of a system change in such a manner that during the change no heat enters or leaves the system is called adiabatic process. Thus in adiabatic process, the total heat of the system remains constant.

Let us consider a gas in a perfectly thermally insulated cylinder fitted with a piston. If the gas is compressed suddenly by moving the piston downward, heat is produced and hence the temperature of the gas will increase. Such a process is adiabatic compression.

If the gas is suddenly expanded by moving the piston outward, energy required to drive the piston is drawn from the internal energy of the gas, causing fall in temperature. This fall in temperature is not compensated by drawing heat from the surroundings. This is adiabatic expansion.

Both the compression and expansion should be sudden, so that there is no time for the exchange of heat. Hence, in an adiabatic process always there is change in temperature.

Expansion of steam in the cylinder of a steam engine, expansion of hot gases in internal combustion engine, bursting of a cycle tube or car tube, propagation of sound waves in a gas are adiabatic processes.

Isothermal Compression

The adiabatic relation between P and V for a gas, is

PV^γ = k, a constant …... (1)

Where γ = specific heat capacity of the gas at constant pressure/specific heat capacity of the gas at constant volume

From the standard gas equation,

PV = RT

P = RT/V

Substituting the value P in (1),

(RT/V) V^γ = constant

T.V^γ-1 = constant

In adiabatic process Q = constant

∴ ∆Q = 0

Specific heat capacity C = ∆Q/m ∆T

∴ C = 0

Refer this video to know more about on, “Adiabatic Process”.

Work done in an adiabatic expansion

Isothermal Expansion

Consider one mole of an ideal gas enclosed in a cylinder with perfectly non conducting walls and fitted with a perfectly frictionless, non conducting piston.

Let P₁, V₁ and T₁ be the initial pressure, volume and temperature of the gas. If A is the area of cross section of the piston, then force exerted by the gas on the piston is

F = P × A, where P is pressure of the gas at any instant during expansion. If we assume that pressure of the gas remains constant during an infinitesimally small outward displacement dx of the piston,

then work done dW = F × dx = P × A dx dW = P dV

Total work done by the gas in adiabatic expansion from volume V₁ to V₂ is

$W = \int_{V_{1}}^{V_{2}}pdV$

But, PV^γ= Constant (k) for adiabatic process

Where, γ = C_P/C_v

So,

$W = \int_{V_{1}}^{V_{2}}kV^{\gamma }dV$

= (k /1- γ)[V₂^1-γ – V₁^1-γ]

= 1/1- γ[kV₂^1-γ – kV₁^1-γ] …... (1)

But, P₂V₂^γ = P₁V₁^γ = k …... (2)

Substituting the value of k in (1),

So, W = (1 /1- γ) [ P₂V₂ – P₁V₁] …... (3)

If T₂ is the final temperature of the gas in adiabatic expansion, then

P₁V₁ = RT₁, P₂V₂ = RT₂

Substituting in equation (3), we get,

W = (1 /1- γ) [ RT₂ – RT₁]

W = (R/1- γ) [T₂ – T₁] …... (4)

This is the equation for the work done during adiabatic process.

Adiabatic processes can occur if the container of the system has thermally-insulated walls or the process happens in an extremely short time.
For an adiabatically expanding ideal monatomic gas which does work on its environment (W is positive), internal energy of the gas should decrease.
In a sense, isothermal process can be considered as the opposite extreme of adiabatic process. In isothermal processes, heat exchange is slow enough so that the system's temperature remains constant.
A system that expands under adiabatic conditions does positive work, so the internal eneregy decreases.
A system that contracts under adiabatic conditions does negative work, so the internal energy increases.
A cyclic transformation whose only final result is to transform heat extracted from a source which is at the same temperature throughout into work is impossible.
A cyclic transformation whose only final result is to transfer heat from a body at a given temperature to a body at a higher temperature is impossible.

Problem (JEE Advanced):

RMS speed of a monoatomic gas is increased by 2 times. If the process is done adiabatically then the ratio of initial volume to final volume will be,

(a) 4 (b) (4)^3/2

Solution:

We know that v_rms∝√T

v_rmsis increased by 2 times, it means its temperature is increased by 4 times.

In adiabatic process,

TV^γ-1 = constant

So, (V_i/V_f) = (T_f/T_i) ^1/γ-1

For monoatomic gas, γ = 5/3

Thus, V_i/V_f = (4)^3/2 = 8

From the above observation, we conclude that, the correct option is (d).

Question 1

In a carnot cycle, the working medium receives heat at a ……… temperature.
a) lower
b) higher
c) constant
d) none of the mentioned

Question 1

In a carnot cycle, the working medium rejects heat at a ……… temperature.
a) lower
b) higher
c) constant
d) none of the mentioned

Question 1

For a given temperature T₁, as the difference between T₁ and T₂increases, the COP of a carnot heat pump
a) increases
b) decreases
c) first increases, then decreases
d) none of the mentioned

Question 1

In a carnot cycle, the working fluid is
a) a real gas
b) an ideal gas
c) a natural gas
d) none of the mentioned

Question 1

The adiabatic process of a carnot cycle needs very ……… motion to complete the adiabatic process.
a) slow
b) fast
c) medium
d) none of the mentioned

Q.1	Q.2	Q.3	Q.4	Q.5
b	b	a	b	b

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