# Chapter Notes: Thermal Expansion and Kinetic Theory of Gases Physics Class 11

Notes for Kinetic theory of Gases chapter of class 11 physics. Dronstudy provides free comprehensive chapterwise class 11 physics notes with proper images & diagram.

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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 $\alpha$ equal to the fraction of the initial length by which a body taken at 0oC has elongated as a result of heating it by 1oC (or by 1 K):

$\alpha = \left( {{l_t}-{l_o}} \right)/{l_o}t$

where lÂ­o is the initial length at 0oC and lt is the length at a temperature t. From this expansion, we can find

${l_t} = {l_o}(1 + \alpha t)$

The dimensions of $\alpha$ are K-1Â  (orÂ Â  oC-1).
The coefficient of cubic expansion is the quantity $\gamma$ equal to the fraction of the initial volume by which the volume of a body taken at 0oC has increased upon heating it by 1oC (or by 1 K):

$\gamma = \left( {{V_t}-{V_o}} \right)/{V_o}t$,

where Vo is the volume of a body at 0oC and Vt is its volume at a temperature t. From this equation, we obtain

${V_t} = {V_o}(1{\rm{ }} + \gamma t)$

The quantity $\gamma$ has also the dimensions of K-1 (orÂ  oC-1).
The coefficient of cubic expansion is about three times larger than the coefficient of linear expansion:

$\gamma = 3\alpha$

The coefficient of cubic expansion $\gamma$ 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 oC. FromÂ 0 oC to 4 oC, water contracts rather than expands. At 4 oC, 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 oC even after the upper layer has been frozen.

Example 1

The lengths l1i = 100 m of iron wire and l1c = 100 m of copper wire are marked off at t1 = 20 oC. What is the difference in lengths of the wires at t2 = 60 oC? The coefficients of linear expansion for iron and copper are ${\alpha _1}$Â = 1.2 xÂ 10-5 K-1 and ${\alpha _c}$Â = 1.7 xÂ 10-5 K-1.

Solution:

${l_{1i}} = {l_o}(1{\rm{ }} + {\alpha _i}{t_1})$Â andÂ  ${l_{2i}} = {l_{oi}}(1{\rm{ }} + {\alpha _i}{t_2})$

The elongation of the iron wires is

${l_{2i}}-{l_{1i}} = {l_{oi}}{\alpha _i}\left( {{t_2}-{t_1}} \right)$.

Substituting Â ${l_{oi}} = {l_{1i}}/\left( {1{\rm{ }} + {f_i}{t_1}} \right)$, we find the elongation of the iron and copper wires

${l_{2i}}-{l_{1i}} = {l_{1i}}{\alpha _i}\left( {{t_2}-{t_1}} \right)/(1{\rm{ }} + {\alpha _i}{t_1})$Â  Â  Â  Â  Â  Â  Â (1)
${l_{2c}}-{l_{1c}} = {l_{1c}}{\alpha _c}\left( {{t_2}-{t_1}} \right)/(1{\rm{ }} + {\alpha _c}{t_1})$ Â  Â  Â  Â  Â  Â  (2)

Subtracting (1) from (2) and considering that ${l_{1i}} = {l_{1c}} = {l_1}$, we obtain

${l_{2c}}-{l_{2i}} = {l_1}{{\left( {{\alpha _c} - {\alpha _i}} \right)\left( {{t_2} - {t_1}} \right)} \over {\left( {1 + {\alpha _c}{t_1}} \right)\left( {1 - {\alpha _i}{t_1}} \right)}} = 19.9mm$

For low values of temperature t, when $\alpha t < 1$, it is not necessary to reduce l1 and l2 to lo1 and lÂ­o2 at t = 0 oC. To a sufficiently high degree of accuracy, we an assume that $\Delta l = l\alpha \Delta t$. Under this assumption, the problem can be solved in a simpler way:

$\Delta {l_i} = {l_{1i}}{\alpha _i}\left( {{t_2}-{t_1}} \right),\Delta {l_c} = {l_{1c}}{\alpha _c}\left( {{t_2}-{t_1}} \right)$

Consequently, since ${l_{1i}} = {l_{1c}} = {l_1}$, we have

$\Delta l = \Delta {l_c} - \Delta {l_i} = {l_1}\left( {{t_2}-{t_1}} \right)({\alpha _c} - {\alpha _i}) = 20{\rm{ }}mm$

It can been seen that the deviation from a more exact value of 19.9 mm amounts to 0.1 mm, i.e. the relative error $\Delta = {\rm{ }}0.1/19.9{\rm{ }} = {\rm{ }}0.5\%$.

Example 2

A solid body floats in a liquid at a temperature t = 0o C and is completely submerged in it at 50o C. What fraction d of volume of the body is submerged in the liquid at 0o C ifÂ  ${\gamma _s}$Â = 0.3 xÂ 10-5 K-2 and of the liquid, ${\gamma _1}$Â = 8.0 xÂ 10-5 K-1?

Solution:

In both the cases the weight of the body will be balanced by the force of buoyancy on it.

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â  AtÂ  to = 0 oC, the buoyancy is ${F_b} = \delta {V_o}{\rho _o}g$ Â  Â  Â  Â  Â  Â (1)

where Vo is the volume of the body and ${\rho _o}$Â is the density of the liquid at to = 0oC. At t = 50 oC, the volume of the body becomes $V = {V_o}(1{\rm{ }} + {\gamma _s}t)$Â and the density of the liquid is ${\rho _1} = {\rho _o}/(1{\rm{ }} + {\gamma _1}t)$. The buoyancy in this case is

${F_b} = {V_o}{\rho _o}g(1{\rm{ }} + {\gamma _s}t)/(1{\rm{ }} + {\gamma _1}t)$Â  Â  Â  Â  Â  Â  Â  Â  (2)

Equating the right-hand sides of equation (1) and (2), we get

$\delta = {\rm{ }}(1{\rm{ }} + {\gamma _s}t)/(1{\rm{ }} + {\gamma _1}t){\rm{ }} = {\rm{ }}96\%$

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