Question #190338

An ideal gas thermometer and a mercury thermometer are calibrated at 0 oC and at 100 oC. The thermal expansion coefficient for mercury is š›¼ = 1 š‘‰0 ( šœ•š‘‰ šœ•š‘‡ ) š‘ƒ = 1.817š‘„10āˆ’4 + 5.90š‘„10āˆ’9šœƒ + 3.45š‘„10āˆ’10šœƒ2 , where šœƒ is the value of the Celsius temperature and š‘‰0 = š‘‰ at šœƒ = 0. What temperature would appear on the mercury scale when the ideal gas scale reads 50 oC? 


Expert's answer

Given,

Ti=0∘CT_i=0^\circ C

Tf=100∘CT_f=100 ^\circ C

α=1Vo(āˆ‚Vāˆ‚T)P=1.817Ɨ10āˆ’4+5.90Ɨ10āˆ’9Īø+3.45Ɨ10āˆ’10Ć—Īø2\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}


⇒α=1Vo(āˆ‚Vāˆ‚T)P\Rightarrow \alpha = \frac{1}{V_o}(\frac{\partial V}{\partial T})_P

ā‡’āˆ‚VVo=Ī±āˆ‚T\Rightarrow \frac{\partial V}{V_o}=\alpha \partial T

Now, taking the integration of both side,

āˆ«āˆ‚VVo=∫(1.817Ɨ10āˆ’4+5.90Ɨ10āˆ’9Īø+3.45Ɨ10āˆ’10Ć—Īø2)āˆ‚T\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}

⇒ln⁔(Vo)=1.817Ɨ10āˆ’4T+5.90Ɨ10āˆ’9T22+3.45Ɨ10āˆ’10T33\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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