91Ó°ÊÓ

According to Charles’ law, the volume of an ideal gas and the absolute temperature are proportional to each other under constant pressure.

The initial- and final-volume circumstances, as well as the final temperature of a carbon monoxide gas sample, are given.

The law of Charles: When the pressure and mass of gas remain constant, Charles’law defines the connection between the volume and temperature.

The volume is proportional to the absolute temperature, according to this formula.

\(\begin{aligned}{}{\rm{V\mu T}}\\\frac{{\rm{V}}}{{\rm{T}}}{\rm{ = constant}}\\{\rm{ }}\frac{{{{\rm{V}}_{\rm{1}}}}}{{{{\rm{T}}_{\rm{1}}}}}{\rm{ = constant ,}}\\\frac{{{{\rm{V}}_{\rm{2}}}}}{{{{\rm{T}}_{\rm{2}}}}}{\rm{ = constant }}\end{aligned}\).

So, \(\;\;\;\frac{{{{\rm{V}}_{\rm{1}}}}}{{{{\rm{T}}_{\rm{1}}}}}{\rm{ = }}\frac{{{{\rm{V}}_{\rm{2}}}}}{{{{\rm{T}}_{\rm{2}}}}}\),

where

\({{\rm{V}}_{\rm{1}}}{\rm{,}}{{\rm{V}}_{\rm{2}}}{\rm{ = }}\)theinitial and final volumes, and

\({{\rm{T}}_{\rm{1}}}{\rm{,}}{{\rm{T}}_{\rm{2}}}{\rm{ = }}\)the initial and final absolute temperatures.

Considering that temperatureismeasured in degrees celsiusrather than kelvins (absolute scale),

we must convert the temperature from degrees celsiusto kelvins.

We have, \({{\rm{0}}^{\rm{^\circ }}}{\rm{ = (0) + 273}}{\rm{.15K}}\)So,

\(\begin{aligned}{}{\rm{55}}{\rm{.}}{{\rm{5}}^{\rm{^\circ }}}{\rm{ = (55}}{\rm{.5) + 273}}{\rm{.15K}}\\{\rm{ = 328}}{\rm{.65K}}\end{aligned}\).

So, the following hold:

Initial volume \(\left( {{{\rm{V}}_{\rm{1}}}} \right){\rm{ = 11}}{\rm{.2\;L}}\).

Final volume \(\left( {{{\rm{V}}_{\rm{2}}}} \right){\rm{ = 13}}{\rm{.3\;L}}{\rm{.}}\)

Initial temperature \(\left( {{{\rm{T}}_{\rm{1}}}} \right){\rm{ = }}\)?

Final temperature \(\left( {{{\rm{T}}_{\rm{2}}}} \right){\rm{ = 328}}{\rm{.65\;K}}{\rm{.}}\)

\(\begin{aligned}{}\frac{{{{\rm{V}}_{\rm{1}}}}}{{{{\rm{T}}_{\rm{1}}}}}{\rm{ = }}\frac{{{{\rm{V}}_{\rm{2}}}}}{{{{\rm{T}}_{\rm{2}}}}}\\{{\rm{T}}_{\rm{1}}}{\rm{ = }}\frac{{{{\rm{V}}_{\rm{1}}}{\rm{ \times }}{{\rm{T}}_{\rm{2}}}}}{{{{\rm{V}}_{\rm{2}}}}}\end{aligned}\)

Initial temperature \(\begin{aligned}{}{T_1} &= \frac{{(11.2\;{\rm{L}}) \cdot (328.65\;{\rm{K}})}}{{13.3\;{\rm{L}}}}\\ &= \frac{{(11.2) \cdot (328.65)}}{{13.3}}\;{\rm{K}}.\\ = \frac{{(3680.88)}}{{13.3}}\;{\rm{K}}.\\ &= 276.75\;{\rm{K}}.\\\\ &= {(276.75 - 273.15)^\circ }.\\ &= {3.6^\circ }\end{aligned}\)

The initial temperature is \({{\rm{T}}_{\rm{1}}}{\rm{ = 276}}{\rm{.75\;K}}\) (or) \({\rm{3}}{\rm{.}}{{\rm{6}}^{\rm{^\circ }}}{\rm{.}}\)