## Thermal Expansion

Experiments show that most of bodies increase their Volume upon heating. The extent of expansion of various bodies is characterized by the temperature coefficient of expansion, or simply the coefficient of expansion. While considering solid which retain their shape during temperature variations, the distinction is made between (a) a change in their linear dimensions (viz. the dimensions in a certain direction), i.e. linear expansion, and (b) a change in the volume of a body, i.e. cubic expansion.

The coefficient of linear expansion is the quantity α equal to the fraction of the initial length by which a body taken at 0^{o}C has elongated as a result of heating it by 1^{o}C (or by 1 K):

α = (*l*_{t} – *l*_{o})/*l*_{o}*t*

where *l*_{o} is the initial length at 0^{o}C and *l*_{t} is the length at a temperature *t*. From this expansion, we can find

*l*_{t} = *l*_{o}(1 + α*t*)

The dimensions of α are K^{-1} (or ^{o}C^{-1}).

The coefficient of cubic expansion is the quantity γ equal to the fraction of the initial volume by which the volume of a body taken at 0^{o}C has increased upon heating it by 1^{o}C (or by 1 K):

γ = (*V _{t}*–

*V*)/

_{o}*V*,

_{o}twhere *V _{o}*is the volume of a body at 0

^{o}C and

*V*is its volume at a temperature

_{t}*t*. From this equation, we obtain

*V _{t}*=

*V*(1 + g

_{o}*t*)

The quantity γ has also the dimensions of K^{-1} (or ^{o}C^{-1}).

The coefficient of cubic expansion is about three times larger than the coefficient of linear expansion:

γ = 3α

The coefficients of γ of cubic expansion for liquids are somewhat higher than for solid bodies, ranging between 10^{-3} and 10^{-4} K^{-1}.

What obeys the general laws of thermal expansion only at a temperature above 4 ^{o}C. From 0 ^{o}C to 4 ^{o}C, water contracts rather than expands. At 4 ^{o}C, water occupies the smallest volume, i.e. it has the highest density. At the bottom of deep lakes, there is denser water in winter, which remains the temperature of 4 ^{o}C even after the upper layer has been frozen.

**Example 10.1**

The lengths l_{1i} = 100 m of iron wire and l_{1c} = 100 m of copper wire are marked off at t_{1} = 20 ^{o}C. What is the difference in lengths of the wires at t_{2} = 60 ^{o}C? The coefficients of linear expansion for iron and copper are α_{1} = 1.2 ´ 10^{-5} K^{-1} and α_{c } = 1.7 ´ 10^{-5} K^{-1}.

** **

*Solution*

* *

**GAS LAWS**

The Gases such as hydrogen, oxygen and helium etc. which can not be liquified easily are called ** permanent gases**. The gases whose molecules are point masses (mass without having volume) and do not attract each other are called

**. Assuming permanent gases to be ideal, through experiment it has been observed that, these gases irrespective of their nature obey the following laws.**

*ideal or perfect gases*** **

**(a) Boyle’s Law **

** **

For a given mass of an Ideal Gas at constant temperature, *the volume of a gas is inversely proportional to its pressure. i.e. *

**(b) Charle’s law **

** **

For a given mass of an ideal gas at constant Pressure, *volume of a gas is directly proportional to its absolute temperature*. i.e.

**(c) Gay – Lussac’s Law **

For a given mass of an ideal gas at constant volume, *pressure of a gas is directly proportional to its absolute temperature*

i.e. P ∝ T

**(d) Avagadro’s Law **

*At the same temperature, pressure and volume all gases contain equal number of molecules*. At STP or NTP (0^{o}C and 1 atm i.e. 273 K and 1.01 × 10^{5} N/m^{2}).

1 mole of an ideal gas º *N _{A}* ( = 6.02 × 10

^{23}) molecules of gas

=22.4 litre of the gas = *M *(molecular weight) gram of the gas

**(e) Graham’s Law **

At constant temperature and pressure, *the rate of diffusion of a gas is inversely proportional to the square root of its density *

**(f) Ideal gas Equation **

Combining first four laws (i.e. Boyle’s law, Charle’s Law, Gay – Lussac’s Law and Avagadro’s law) we get one single equation for an ideal gas, i.e.

**(g) Dalton’s Law **

The pressure exerted by a gaseous mixture is equal to the sum of partial pressure of each component present in the mixture.

*i.e. P = P _{1} + P_{2} + …………*

**(10.2)**

#### Example: 10.2

A vessel of volume 2 × 10^{-3} m^{3} contains 0.1 mol of *hydrogen *gas and 0.2 mol of *helium*. If the temperature of the mixture is 300 K, calculate the pressure due to component gases and the mixture. (*R* = 8.31 J/mol K)

*Solution *

The post Thermal Expansion & Different Gas Laws appeared first on Quantemporary.

*This post first appeared on Articles Of Physics Topics For IIT-JEE, PMT, IB, SAT, AP Students, please read the originial post: here*