Question 1.42 - Chemistry: Molecular Nature Of Matter And Change

Question

Perform the following conversions:

$$\begin{gathered} \mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{68}\mathbf{ }\mathbf{°}\mathbf{F}\mathbf{ }\mathbf{(}\mathbf{a}\mathbf{ }\mathbf{pleasant}\mathbf{ }\mathbf{spring}\mathbf{ }\mathbf{day}\mathbf{)}\mathbf{ }\mathbf{to}\mathbf{ }\mathbf{°}\mathbf{C}\mathbf{ }\mathbf{and}\mathbf{ }\mathbf{K}\mathbf{ } \\ \mathbf{\left(}\mathbf{b}\mathbf{\right)}\mathbf{ }\mathbf{ }\mathbf{164}\mathbf{°}\mathbf{C}\mathbf{ }\mathbf{(}\mathbf{the}\mathbf{ }\mathbf{boiling}\mathbf{ }\mathbf{point}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{methane}\mathbf{,}\mathbf{ }\mathbf{the}\mathbf{ }\mathbf{main}\mathbf{ }\mathbf{component}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{natural}\mathbf{ }\mathbf{gas}\mathbf{)}\mathbf{ }\mathbf{to}\mathbf{ }\mathbf{K}\mathbf{ }\mathbf{and}\mathbf{ }\mathbf{°}\mathbf{F}\mathbf{ } \\ \mathbf{\left(}\mathbf{c}\mathbf{\right)}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{0}\mathbf{ }\mathbf{K}\mathbf{ }\mathbf{(}\mathbf{absolute}\mathbf{ }\mathbf{zero}\mathbf{,}\mathbf{ }\mathbf{theoretically}\mathbf{ }\mathbf{the}\mathbf{ }\mathbf{coldest}\mathbf{ }\mathbf{possible}\mathbf{ }\mathbf{temperature}\mathbf{)}\mathbf{ }\mathbf{to}\mathbf{ }\mathbf{°}\mathbf{C}\mathbf{ }\mathbf{and}\mathbf{ }\mathbf{°}\mathbf{F} \end{gathered}$$

Step-by-Step Solution

Answer

Answer

Temperature is defined in three scales, and values in one scale are interconvertible to another scale. The Kelvin and Celsius scales only differ in zero-point. Hence they can be converted in one step while the Fahrenheit scale differs from the Celsius scale in unit size and zero-point both. So for converting a temperature in Celsius to Fahrenheit, two steps are required.

$$\begin{gathered} \begin{array}{rcl}& & 68°\mathrm{F} = 20°\mathrm{C} \\ = 293.15 \mathrm{K} \\ 164°\mathrm{C} = 327.2°\mathrm{F} \\ = 437.15\mathrm{K} \\ 0 \mathrm{K} = -273.15°\mathrm{C} \\ = -459.67°\mathrm{F}\end{array} \end{gathered}$$

Relation between Three Scales of Temperature

$$\begin{gathered} \mathbf{The}\mathbf{ }\mathbf{temperature}\mathbf{ }\mathbf{in}\mathbf{ }\mathbf{the}\mathbf{ }\mathbf{Celsius}\mathbf{ }\mathbf{and}\mathbf{ }\mathbf{Kelvin}\mathbf{ }\mathbf{scales}\mathbf{ }\mathbf{is}\mathbf{ }\mathbf{related}\mathbf{ }\mathbf{as}\mathbf{ }\mathbf{the}\mathbf{ }\mathbf{given}\mathbf{ }\mathbf{equation}\mathbf{:} \\ \mathbf{T}\mathbf{ }\mathbf{(}\mathbf{in}\mathbf{ }\mathbf{K}\mathbf{)}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{T}\mathbf{ }\mathbf{(}\mathbf{in}\mathbf{ }\mathbf{°}\mathbf{C}\mathbf{)}\mathbf{ }\mathbf{+}\mathbf{ }\mathbf{273}\mathbf{.}\mathbf{15}\mathbf{ }\mathbf{K} \\ \mathbf{The}\mathbf{ }\mathbf{temperature}\mathbf{ }\mathbf{in}\mathbf{ }\mathbf{Celsius}\mathbf{ }\mathbf{and}\mathbf{ }\mathbf{Fahrenheit}\mathbf{ }\mathbf{are}\mathbf{ }\mathbf{related}\mathbf{ }\mathbf{as}\mathbf{ }\mathbf{the}\mathbf{ }\mathbf{given}\mathbf{ }\mathbf{equation}\mathbf{:} \\ \mathbf{T}\mathbf{ }\mathbf{(}\mathbf{in}\mathbf{ }\mathbf{°}\mathbf{F}\mathbf{)}\mathbf{ }\mathbf{=}\mathbf{ }\frac{\mathbf{9}}{\mathbf{5}}\mathbf{T}\mathbf{ }\mathbf{(}\mathbf{in}\mathbf{ }\mathbf{°}\mathbf{C}\mathbf{)}\mathbf{ }\mathbf{+}\mathbf{ }\mathbf{32}\mathbf{ }\mathbf{ }\mathbf{ } \end{gathered}$$

Conversion of 68 °F (a pleasant spring day) to °C and K

As we have seen, 

$\begin{array}{rcl}\mathrm{T} (\mathrm{in} °\mathrm{F}) & =& \frac{9}{5} \mathrm{T}(\mathrm{in} °\mathrm{C}) + 32\\ & & \mathrm{So},\\ \mathrm{T} (°\mathrm{C}) & =& \frac{5}{9}\left[\mathrm{T} (°\mathrm{F}) - 32\right]\\ & =& \frac{5}{9}\left[68 - 32\right]\\ & =& 20°\mathrm{C}\end{array}$

And 

$\begin{array}{rcl}\mathrm{T} (\mathrm{in} \mathrm{K}) & =& \mathrm{T}(°\mathrm{C}) + 273.15\\ & =& 20 + 273.15\\ & =& 293.15 \mathrm{K}\\ & & \mathrm{Hence},\\ 68°\mathrm{F} & =& 20°\mathrm{C}\\ & =& 293.15 \mathrm{K}\end{array}$

164 °C (the boiling point of methane) to K and °F

First, convert Celsius to Kelvin using the formula given in Step 1.

$\begin{array}{rcl}T \left(K\right)& =& T(°C) + 273.15\\ & =& 164 + 273.15\\ & =& 437.15 K\\ & & \end{array}$

Now, from Celsius to Fahrenheit:

$\begin{array}{rcl}T(°F) & =& \frac{9}{5}T(°C) +32\\ & =& \frac{9}{5}\left(164\right) + 32\\ & =& 327.2°F\end{array}$

Hence,

$\begin{array}{rcl}164°C& =& 327.2°F\\ & =& 437.15 K\end{array}$

0 K (absolute zero) to °C and °F

First, convert Kelvin to Celsius using the formula given in Step 1.

$\begin{array}{rcl}T (°C) & =& T\left(K\right) -273.15\\ & =& 0 - 273.15\\ & =& -273.15°C\end{array}$

Now, from Celsius to Fahrenheit:

$\begin{array}{rcl}T(°F) & =& \frac{9}{5}T(°C) + 32\\ & =& \frac{9}{5}(-273.15) + 32\\ & =& -459.67°F\end{array}$

Hence,

$\begin{array}{rcl}0 K & =& -273.15°C\\ & =& -459.67°F\end{array}$