Thermal expansion – ravnik.eu

Problem

Mercury in a glass thermomether expands to encompass the whole glass volume at $50^\circ C$ . For how much does the pressure in mercury rise, when the temperature is raised from $50^\circ C$ to $51^\circ C$ ?

Assumptions:
– Compressibility of mercury is $\chi=4\cdot10^{-6}~bar^{-1}$
– Glass is rigid, i.e. it does not expand

Cases:
a) Thermal expansion coefficient of mercury is temperature independent at $\beta_0=1.8\cdot 10^{-4}~K^{-1}$
b) Thermal expansion coefficient of mercury is $\beta_0=1.8\cdot 10^{-4}~K^{-1}$ at $50^\circ C$ and increases by $1\%$ when the temperature is raised to $51^\circ C$ .

Data

$T_0=50^\circ C$
$T_1=51^\circ C$
$\chi=4\cdot10^{-6}~bar^{-1}$
$\beta_0=1.8\cdot 10^{-4}~K^{-1}$
$\beta_1=1.01\beta_0$
$\Delta p =?$

Solution, case a)

In a liquid, such as mercury the equation relating its volume with pressure and temperature is

$\frac{dV}{V}=\beta dT-\chi dp.$

Since the glass is rigid, its volume does not change, so $dV/V=0$ . Simplifying we obtain a direct relationship between a change in temperature and a change in pressure:

$\beta dT=\chi dp.$

When a liquid is heated from $T_1$ to $T_2$ we expect the pressure to rise from $p_1$ to $p_2$ . As in this case both thermal expansion and compressibilty are independent of temperature and pressure, we can simply state

$\beta(T_1-T_0)=\chi(p_1-p_0).$

Finally, the pressure increase is

$\Delta p = p_1-p_0=\frac{\beta_0(T_1-T_0)}{\chi}=45~bar.$

Solution, case b)

In this case, we have to take into account that the thermal expansion coefficient increases with temperature. It is $\beta_0$ at $T_0$ and $\beta_1$ at $T_1$ . We do not know, how it increases, but since the temperature difference is small, we can assume, that it increases linearly. Assuming linear dependence, we write an equation of a line:

$\beta(T)=\beta_0+\frac{\beta_1-\beta_0}{T_1-T_0}(T-T_0).$

We discovered in case a) that the relationship between a change in temperature and a change in pressure is:

$\beta(T) dT=\chi dp.$

As the thermal expansion is now a function of temperature, we should integrate this equation from the state $(T_0,p_0)$ to the final state $(T_1,p_1)$ :

$\int_{T_0}^{T_1}\beta(T) dT=\int_{p_0}^{p_1}\chi dp=\chi\Delta p.$

On the pressure side of the equations, compressiblity is constant, so the integral simply transforms the pressure differential $dp$ to the pressure difference $\Delta p$ . The integral on the left hand side is evaluated as follows:

$\int_{T_0}^{T_1}\beta(T) dT=\int_{T_0}^{T_1}\left(\beta_0+\frac{\beta_1-\beta_0}{T_1-T_0}(T-T_0)\right) dT=$

$=\int_{T_0}^{T_1}\beta_0 dt+\frac{\beta_1-\beta_0}{T_1-T_0}\int_{T_0}^{T_1}TdT -\frac{\beta_1-\beta_0}{T_1-T_0}T_0\int_{T_0}^{T_1}dT=$

$=\beta_0\Delta T+\frac{\beta_1-\beta_0}{T_1-T_0}\frac{T_1^2-T_0^2}{2}-(\beta_1-\beta_0)T_0=$

$=\beta_0\Delta T+(\beta_1-\beta_0)\left(\frac{T_1+T_0}{2}-T_0\right)=$

$=\beta_0\Delta T+(\beta_1-\beta_0)\frac{\Delta T}{2}=$

$=\frac{\beta_1+\beta_0}{2}\Delta T.$

Combining the integral on the left and right hand side we finally obtain

$\Delta p=\frac{\beta_1+\beta_0}{2\chi}\Delta T=45.2~bar.$

Since we assumed linear dependence of thermal expansion with temperature it is not surprising that the solution of case b) is conceptually the same as the solution in case a) but with average thermal expansion $(\beta_1+\beta_0)/2$ . At the same time, we notice, that the error made when neglecting thermal expansion is $0.2~bar$ .

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