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,
Ti=0∘CT_i=0^\circ CTi=0∘C
Tf=100∘CT_f=100 ^\circ CTf=100∘C
α=1Vo(∂V∂T)P=1.817×10−4+5.90×10−9θ+3.45×10−10×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=Vo1(∂T∂V)P=1.817×10−4+5.90×10−9θ+3.45×10−10×θ2
⇒α=1Vo(∂V∂T)P\Rightarrow \alpha = \frac{1}{V_o}(\frac{\partial V}{\partial T})_P⇒α=Vo1(∂T∂V)P
⇒∂VVo=α∂T\Rightarrow \frac{\partial V}{V_o}=\alpha \partial T⇒Vo∂V=α∂T
Now, taking the integration of both side,
∫∂VVo=∫(1.817×10−4+5.90×10−9θ+3.45×10−10×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}∫Vo∂V=∫(1.817×10−4+5.90×10−9θ+3.45×10−10×θ2)∂T
⇒ln(Vo)=1.817×10−4T+5.90×10−9T22+3.45×10−10T33\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}⇒ln(Vo)=1.817×10−4T+25.90×10−9T2+3.45×10−103T3
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