# 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:

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

$$K=\frac{°F-32}{1.8}+273.15$$