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 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? 


1
Expert's answer
2021-05-09T15:12:33-0400

Given,

Ti=0CT_i=0^\circ C

Tf=100CT_f=100 ^\circ C

α=1Vo(VT)P=1.817×104+5.90×109θ+3.45×1010×θ2\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(VT)P\Rightarrow \alpha = \frac{1}{V_o}(\frac{\partial V}{\partial T})_P

VVo=αT\Rightarrow \frac{\partial V}{V_o}=\alpha \partial T

Now, taking the integration of both side,

VVo=(1.817×104+5.90×109θ+3.45×1010×θ2)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)=1.817×104T+5.90×109T22+3.45×1010T33\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}


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS