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?
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}"
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