Question 129

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, $C = \frac{5}{9} (F - 32)$
$\begin{matrix} Gas & Freezing point at sea level (^{o} F) & Freezing point at sea level (^{o} C) Hydrogen & - 435 Krypton & - 251 Oxygen & - 369 Helium & - 458 Argon & - 309 \end{matrix}$

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Solution

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)
$C = \frac{5}{9} (- 309 - 32) = \frac{5}{9} (- 341) = - 189.44 \Rightarrow C = - 189.44 \Rightarrow C = - 189$

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