# Isentropic Process Nitrogen with Pressure Ratio

Nitrogen (k=1.4) is expanded isentropically. Its temperature changes from 620°F to 60°F. Find the Pressure Ratio:

Answer is 12.91

Simply use Isentropic Relation.

$$620°F = (1799/3) K$$ $$60°F = (2597/3) K$$

$$\frac{T_1}{T_2} = \frac{P_1}{P_2}^{\frac{k-1}{1.4}}$$ $$\frac{(1799/3)}{(2597/3)} = \frac{P_1}{P_2}^{\frac{1.4-1}{1.4}}$$ $$\frac{(1799/3)}{(2597/3)}^{\frac{1.4}{1.4-1}} = \frac{P_1}{P_2}$$

I get 0.2766669

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

• Your temperature conversion shows that 620 F = 600 K, and that 60 F = 866 K, which implies 60 F > 620 F. That's not correct, so look there for problems. – Todd Minehardt Jul 7 '15 at 1:42

## 1 Answer

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

$$K=\frac{°F-32}{1.8}+273.15$$

(If you want, double check your conversion using an online conversion calculator such as Metric Conversion).

Then continue your calculations as you have done (I just did and got the correct answer).