Production of Low Temperature, Physics tutorial

Cooling Process:

In general, two processes are involved in cooling a gas that is; isothermal process followed by adiabatic process. For instance, to cool the gas by adiabatic decompression, gas is first compressed isothermally that is at constant temperature. This is performed by compressing gas in the vessel that is not insulated and wait long enough for gas to lose any heat which is produced because of compression. In this way, a constant temperature is kept. Heat that is produced can be explained in terms of law PV/T constant. Then, vessel containing gas is insulated and gas is permitted to expand adiabatically (i.e. no heat transfer is permitted between vessel and surrounding). This type of expansion brings about reduction in temperature. Expression for change in temperature with respect to pressure at constant entropy (∂T/∂P)S can be derived. Process can be repeated until desired temperature is attained. This method is called cooling by adiabatic decompression.

Methods of Cooling:

Few methods being utilized for low temperature cooling are:

• laser cooling
• evaporative cooling (e.g. evaporation of 3He)
• cooling by liquefaction

The procedure of cooling by adiabatic demagnetization has been utilized to get extremely low temperature. Cooling by adiabatic demagnetization is thus explained below.

Different methods have been used to get very low temperature. Method of adiabatic demagnetization has been utilized to get extremely low temperature. Figure given below shows S-T diagram for adiabatic demagnetization. Magnetic field B is zero along curve labeled B = 0 while magnetic field is B along curve labeled B. In process ab, the sample of paramagnetic salt (like cerium magnesium nitrate) already cooled to low temperature by other means (like by contact with the bath of liquid helium), is magnetized isothermally. Sample is frequently suspended in atmosphere of helium that can conduct away any heat that is produced, and therefore keeps process isothermal. Therefore, process ab is isothermal magnetization. Then, in process bc (that is adiabatic demagnetization), paramagnetic salt is insulated (by pumping out the helium) and then demagnetized adiabatically. This procedure of isothermal magnetization followed by the adiabatic Demagnetization can be repeated over and over again until desired temperature is achieved. Temperature close to 0 K have been reached in this manner.

Magnetic dipole moment P of the sample is maximum torque it experiences in unit field B. Torque is provided by τ = P x B. Magnetization M of the specimen is expressed by B = μH = μ0(H + M).

Magnetization is also equal to magnetic moment per unit volume. Differential form of work for the magnetic system is BdM

dW = BdM

BdM is work done per unit volume on the isotropic sample in increasing its magnetization from M to M + dM.

If we add heat to the magnetisable sample, and perform work per unit volume on it by putting it in the magnetic field B and thus increase its magnetization by dM, then, given there is no change in volume, increase in its internal energy per unit volume is provided by

dU = TdS + BdM

In this magnetic context, we can state state functions H, A, and G per unit volume by:

H = U - BM

A = U -TS

G = H - TS = A - BM

And differential forms as,

dH = TdS - MdB

M is dipole moment per unit volume, in NmT-1m-3, which is same as magnetization in Am-1 We can derive the expression for lowering of temperature in the adiabatic decompression (∂T/∂P)s. Also using same argument, step-by-step, for lowering of temperature in the adiabatic demagnetization (∂T/∂B)S

Considering entropy as the function of temperature and pressure (i.e. S(T, P)),

(∂S/∂T)P(∂T/∂P)S(∂P/∂S)T = -1

In the reversible process dS = dQ/T, and in the isobaric process,

dQ = CPdT

I.e. TdS = CPdT

From equation, partial derivative of S with respect to T at constant pressure provides:

(∂S/∂T)P = CP/T

Also from the Maxwell relation

(∂S/∂P)T = - (∂V/∂T)P

After solving the equations we get:

(∂T/∂P)s = T/CP(∂V/∂T)P

If gas is the ideal gas, equation of state is PV = nRT , so that

(∂V/∂T)P = nR/P = V/T

((∂T/∂P)S) = V/CP

Same argument as above can be utilized for adiabatic demagnetization (∂T/∂B)S. We can consider entropy as the function of temperature and magnetic field i.e. S(T, B), we have

(∂S/∂T)B(∂T/∂B)S(∂B/∂S)T = -1

Then

(∂T/∂B)S = - (∂T/∂S)B(∂S/∂B)T

In the reversible process dS = dQ/T, and in the constant magnetic field, dQ = CBdT, CB is the heat capacity per unit volume (i.e. heat required to raise the temperature of the unit volume by one degree)

In the constant magnetic field

(∂S/∂T)B = CB/T

Now for the paramagnetic material, magnetization, for the given filed is proportional to B and it falls off inversely as temperature (that is equation of state).

That is M = aB/T

Further solving (∂T/∂B)S = M/CB

This equation provides cooling effect, i.e. variation of temperature with magnetic field at constant entropy, in terms of magnetization M and heat capacity CB. Cooling effect is particularly effective at low temperature when CB is small.

Entropy and Temperature:

Cooling by adiabatic demagnetization involves successive isothermal magnetizations followed by adiabatic demagnetizations, and this suggests that some insight in process might be attained by following it on the entropy-temperature (ST) diagram. Cooling effect (∂T/∂B)S in process bc and other subsequent processes indicated by horizontal lines linking two curves. The complete cooling is indicated in processes ab and bc that can be repeated until desired temperature is attained.

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