Pressure ratio for isentropic expansion

### Problem



> 
> Nitrogen $(\gamma = 1.4)$ is expanded isentropically. Its temperature changes from $\pu{620 °F}$ to $\pu{60 °F}.$ Find the pressure ratio.
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### Answer



> 
>  $p\_1/p\_2 = 12.91$
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### My approach


$$\pu{620 °F} = (1799/3)~\pu{K}$$
$$\pu{60 °F} = (2597/3)~\pu{K}$$


Using isentropic relation:


$$\frac{T\_1}{T\_2} = \left(\frac{p\_1}{p\_2}\right)^\frac{\gamma-1}{\gamma}$$
$$\frac{1799/3}{2597/3} = \left(\frac{p\_1}{p\_2}\right)^{\frac{1.4-1}{1.4}}$$
$$\frac{p\_1}{p\_2} = \left(\frac{1799/3}{2597/3}\right)^{\frac{1.4}{1.4-1}} = 0.2766669$$


What am I doing wrong? Is the problem set possibly wrong?




As stated by @Todd in the comments, your temperature conversions are incorrect. To convert between °F and K, use the following conversion:


$$\pu{K} = \frac{\pu{°F}-32}{1.8}+273.15$$


If you want, double check your conversion using an online conversion calculator such as [Metric Conversion](http://www.metric-conversions.org/temperature/fahrenheit-to-kelvin.htm).
Then continue your calculations as you have done (I just did and got the correct answer).


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