Question 17.4: If the value of αF at a temperature of T is known to be −2.0......
If the value of α_F at a temperature of T is known to be −2.0 × 10^{−5} per °K, what is the value of α_F at a temperature of 4T?
Since α_{\mathrm{F}}=\frac{{C}}{\sqrt{T}\ln[{1}/{\bf p}_{0}]}, it follows that α_{\mathrm{F}}(4\mathrm{T})={\frac{\mathbf{C}}{\sqrt{4\mathrm{T}}\ln[1/\mathrm{p}_{\mathrm{o}}]}} = {\frac{1/2\mathbf{C}}{\sqrt{\mathbf{T}}\ln[1/\mathbf{p}_{0}]}}, so
\frac{\alpha_{\mathrm{F}}(4\mathrm{T})}{\alpha_{\mathrm{F}}(\mathrm{T})}=\frac{1/2\,{\mathrm{C}}{\sqrt{\mathrm{T}}\,\mathrm{ln}[\,1/{\mathrm{p}}_{\mathrm{o}}]}}{\mathrm{C}\sqrt{\mathrm{T}}\,\mathrm{ln}[\,1/{\mathrm{p}}_{\mathrm{o}}\,]}.
Most of the terms in this expression cancel, and after some additional algebra, we are left with
\frac{\alpha_{\mathrm{F}}(4\mathrm{T})}{\alpha_{\mathrm{F}}(\mathrm{T})} = \frac{1}{2}
Thus if α_F(T) = -2.0 × 10^{-5} per ºK, then α_F(4T) = \frac{1}{2}α_F(T) = -1.0 × 10^{-5} per ºK
So if we increase the fuel temperature by a factor of 4, α_F becomes 50% smaller. Note, however, that the value of α_F is still a negative number in this case.