91Ó°ÊÓ

The force applied perpendicular to an object's surface per unit area across which that force is spread is known as pressure.

When there is a constant amount of gas in a closed and rigid container, Gay-Lussac's law (or the pressure law) states the relationship between the pressure and temperature of the gas.

The absolute pressure is exactly proportional to the temperature, according to Gay-Lussac's law.

\(\begin{aligned}{}{\rm{P}} \propto {\rm{T}}\\\frac{{\rm{P}}}{{\rm{T}}}{\rm{ = constant}}\\\frac{{{{\rm{P}}_{\rm{1}}}}}{{{{\rm{T}}_{\rm{1}}}}}{\rm{ = constant}}\\\frac{{{{\rm{P}}_{\rm{2}}}}}{{{{\rm{T}}_{\rm{2}}}}}{\rm{ = constant}}\\\frac{{{{\rm{P}}_{\rm{1}}}}}{{{{\rm{T}}_{\rm{1}}}}}{\rm{ = }}\frac{{{{\rm{P}}_{\rm{2}}}}}{{{{\rm{T}}_{\rm{2}}}}}\end{aligned}\)

\({{\rm{P}}_{\rm{1}}}\),\({{\rm{P}}_{\rm{2}}}\)are the initial and final pressures, respectively.

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

However, considering that temperatures are measured in degrees Celsius rather than kelvins (absolute scale).

As a result, we must convert temperatures from degrees Celsius to kelvins.

\(\begin{aligned}{}{\rm{0^\circ = (0) + 273}}{\rm{.15K}}\\{\rm{23^\circ = (23) + 273}}{\rm{.15K}}\\{\rm{ = 296}}{\rm{.15K}}\\\\{\rm{475^\circ = (475) + 273}}{\rm{.15K}}\\{\rm{ = 748}}{\rm{.15K}}\end{aligned}\)

The initial pressure (\({{\rm{P}}_{\rm{1}}}\))=\({\rm{1344 torr}}\)

Final Pressure (\({{\rm{P}}_{\rm{2}}}\)) =? torr

Temperature at the start (\({{\rm{T}}_{\rm{1}}}\)) =\({\rm{296}}{\rm{.15K}}\).

The final temperature (\({{\rm{T}}_{\rm{2}}}\)) = \({\rm{748}}{\rm{.15K}}\).

\(\begin{aligned}{}\frac{{{{\rm{P}}_{\rm{1}}}}}{{{{\rm{T}}_{\rm{1}}}}}{\rm{ = }}\frac{{{{\rm{P}}_{\rm{2}}}}}{{{{\rm{T}}_{\rm{2}}}}}\\{{\rm{P}}_{\rm{2}}}{\rm{ = }}\frac{{{{\rm{P}}_{\rm{1}}}{\rm{ \times }}{{\rm{T}}_{\rm{2}}}}}{{{{\rm{T}}_{\rm{1}}}}}\\{\rm{ = }}\frac{{{\rm{1344 torr \times 748}}{\rm{.15K}}}}{{{\rm{296}}{\rm{.15K}}}}\\{\rm{ = 3,395 Torr}}\\{\rm{ = 3}}{\rm{.395 \times 1}}{{\rm{0}}^{\rm{3}}}{\rm{Torr}}\\{\rm{or}}\\{\rm{ = 3}}{\rm{.4 \times 1}}{{\rm{0}}^{\rm{3}}}{\rm{Torr}}\end{aligned}\)

Therefore, final pressure is \({\rm{3}}{\rm{.4 \times 1}}{{\rm{0}}^{\rm{3}}}{\rm{Torr}}\).