A 2.50 L volume of hydrogen measured at-196⁰C is warmed to100⁰C. Calculate the volume of the gas at the higher temperature, assuming no change in pressure.

As per Charles’ law, the volume of an ideal gas and 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’s 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,

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

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

Considering that temperatures are measured in degrees celsiusrather than kelvins (absolute scale),

we must convert temperatures from degrees celsiusto kelvins.

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

So,

\(\begin{aligned}{}{\rm{ - 19}}{{\rm{6}}^{\rm{^\circ }}}{\rm{ = ( - 196) + 273}}{\rm{.15K}}\\{\rm{ = 77}}{\rm{.15K}}\end{aligned}\)

\(\begin{aligned}{}{\rm{10}}{{\rm{0}}^{\rm{^\circ }}}{\rm{ = (100) + 273}}{\rm{.15K}}\\{\rm{ = 373}}{\rm{.15K}}\end{aligned}\).

Initial volume \(\left( {{{\rm{V}}_{\rm{1}}}} \right){\rm{ = 2}}{\rm{.5\;L}}{\rm{.}}\;\;\;\)

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

Initial temperature \(\left( {{{\rm{T}}_{\rm{1}}}} \right){\rm{ = 77}}{\rm{.15\;K}}{\rm{.}}\;\;\;\)

Final temperature \(\left( {{{\rm{T}}_{\rm{2}}}} \right){\rm{ = 373}}{\rm{.15\;K}}{\rm{.}}\)

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

The final volume is

\(\begin{aligned}{}{{\rm{V}}_{\rm{2}}}{\rm{ = }}\frac{{{\rm{(2}}{\rm{.5\;L) \times (373}}{\rm{.15\;K)}}}}{{{\rm{77}}{\rm{.15\;K}}}}\\{\rm{ = }}\frac{{{\rm{(2}}{\rm{.5) \times (373}}{\rm{.15)}}}}{{{\rm{77}}{\rm{.15}}}}{\rm{L}}{\rm{.}}\\{\rm{ = }}\frac{{{\rm{(932}}{\rm{.875)}}}}{{{\rm{77}}{\rm{.15}}}}{\rm{\;L}}{\rm{.}}\\{\rm{ = 12}}{\rm{.09\;L}}{\rm{.}}\end{aligned}\)

Therefore, the final volume of the gas at the higher temperature is \({{\rm{V}}_{\rm{2}}}{\rm{ = 12}}{\rm{.09\;L}}\).