Question

The given table shows the freezing points in ${}^{o}F$ of different gases at sea level. Convert each of these into ${}^{o}C$ to the nearest integral value using the relation and complete the table, $C=\frac{5}{9}(F-32)$
$\begin{array}{ccc}& & \\ \text{Gas}& \text{Freezing point at sea level}{(}^{o}F)& \text{Freezing point at sea level}{(}^{o}C)\\ \text{Hydrogen}& -435& \\ \text{Krypton}& -251& \\ \text{Oxygen}& -369& \\ \text{Helium}& -458& \\ \text{Argon}& -309& \end{array}$

Answer 1

We have, $C=\frac{5}{9}(F-32)$
For Hydrogen, put F= (-435)
$\Rightarrow C=\frac{5}{9}(-435-32)=\frac{5}{9}\times (-467)=-259.44\Rightarrow C=-259.44\Rightarrow C=-259$
For Krypton, put F= (-251)
$\Rightarrow C=\frac{5}{9}(-251-32)=\frac{5}{9}\times (-283)=-157.22\Rightarrow C=-157.22\Rightarrow C=-157$
For Oxygen, put F = (-369)
$C=\frac{5}{9}(-369-32)=\frac{5}{9}\times (-401)\Rightarrow C=222.7\Rightarrow C=223$
For Helium, put F = (-458)
$C=\frac{5}{9}(-458-32)=\frac{5}{9}\times (-490)=-272.22\Rightarrow C=-272.22\Rightarrow C=-272$
For Argon, put F = (-309)
$C=\frac{5}{9}(-309-32)=\frac{5}{9}(-341)=-189.44\Rightarrow C=-189.44\Rightarrow C=-189$