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The triple points of neon and carbon dioxide are $24.57$ $K$ and $216.55 \mathrm{~K}$ respectively. Express these temperatures on the Celsius and Fahrenheit scales.
Solution:
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Verified Answer
If temperature in Kelvin scale is $\mathrm{K}$ and that in Celsius scale is $\mathrm{C}$ then, $\mathrm{C}=\mathrm{K}-273.15$
$$
\begin{gathered}
\therefore \quad \text { For Neon, } \mathrm{C}=24.57-273.15=-248.58^{\circ} \mathrm{C} ; \text { For } \mathrm{CO}_2 \text {, } \\
\therefore \mathrm{C}=216.55-273.15=-56.60^{\circ} \mathrm{C}
\end{gathered}
$$
If temperature in Fahrenheit scale is $\mathrm{F}$ then,
$$
\begin{aligned}
&\begin{aligned}
\frac{\mathrm{F}-32}{180} &=\frac{\mathrm{K}-273.15}{100}
\end{aligned} \\
&\text { For neon, } \begin{aligned}
\mathrm{F} &=\frac{180}{100}(24.57-273.15)+32 \\
&=-415.44^{\circ} \mathrm{F}
\end{aligned}
\end{aligned}
$$
For $\mathrm{CO}_2, \mathrm{~F}=\frac{180}{100}(216.55-273.15)+32$
$$
=-69.88^{\circ} \mathrm{F}
$$
$$
\begin{gathered}
\therefore \quad \text { For Neon, } \mathrm{C}=24.57-273.15=-248.58^{\circ} \mathrm{C} ; \text { For } \mathrm{CO}_2 \text {, } \\
\therefore \mathrm{C}=216.55-273.15=-56.60^{\circ} \mathrm{C}
\end{gathered}
$$
If temperature in Fahrenheit scale is $\mathrm{F}$ then,
$$
\begin{aligned}
&\begin{aligned}
\frac{\mathrm{F}-32}{180} &=\frac{\mathrm{K}-273.15}{100}
\end{aligned} \\
&\text { For neon, } \begin{aligned}
\mathrm{F} &=\frac{180}{100}(24.57-273.15)+32 \\
&=-415.44^{\circ} \mathrm{F}
\end{aligned}
\end{aligned}
$$
For $\mathrm{CO}_2, \mathrm{~F}=\frac{180}{100}(216.55-273.15)+32$
$$
=-69.88^{\circ} \mathrm{F}
$$
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