Answer to Question #190338 in Molecular Physics | Thermodynamics for wilfreed

Question

An ideal gas thermometer and a mercury thermometer are calibrated at 0 oC and at 100 oC. The thermal expansion coefficient for mercury is 𝛼 = 1 𝑉0 ( 𝜕𝑉 𝜕𝑇 ) 𝑃 = 1.817𝑥10−4 + 5.90𝑥10−9𝜃 + 3.45𝑥10−10𝜃² , where 𝜃 is the value of the Celsius temperature and 𝑉0 = 𝑉 at 𝜃 = 0. What temperature would appear on the mercury scale when the ideal gas scale reads 50 oC? 

Answer 1

Given,

$T_i=0^\circ C$

$T_f=100 ^\circ C$

$\alpha = \frac{1}{V_o}(\frac{\partial V}{\partial T})_P =1.817\times 10^{-4}+5.90 \times 10^{-9}\theta +3.45\times 10^{-10}\times{\theta^2}$

$\Rightarrow \alpha = \frac{1}{V_o}(\frac{\partial V}{\partial T})_P$

$\Rightarrow \frac{\partial V}{V_o}=\alpha \partial T$

Now, taking the integration of both side,

$\int\frac{\partial V}{V_o}=\int(1.817\times 10^{-4}+5.90 \times 10^{-9}\theta +3.45\times 10^{-10}\times{\theta^2}){\partial T}$

$\Rightarrow \ln(V_o)=1.817\times 10^{-4}T+\frac{5.90\times 10^{-9} T^2}{2}+3.45\times 10^{-10}\frac{T^3}{3}$