Use Charles' Law gas equation to find the temperature:

$\overline{)\frac{{\mathbf{P}}_{\mathbf{1}}}{{\mathbf{T}}_{\mathbf{1}}}{\mathbf{=}}\frac{{\mathbf{P}}_{\mathbf{2}}}{{\mathbf{T}}_{\mathbf{2}}}}\phantom{\rule{0ex}{0ex}}{\mathbf{T}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{(}\mathbf{24}\mathbf{+}\mathbf{273}\mathbf{.}\mathbf{15}\mathbf{}\mathbf{K}\mathbf{)}\mathbf{(}\mathbf{155}\mathbf{}\overline{)\mathbf{atm}}\mathbf{)}}{\mathbf{16}\mathbf{,}\mathbf{500}\mathbf{}\overline{)\mathbf{kPa}}\mathbf{}\mathbf{\left(}{\displaystyle \frac{1\overline{)\mathrm{Pa}}}{1x{10}^{-3}\overline{)\mathrm{kPa}}}}\mathbf{\right)}\mathbf{\left(}{\displaystyle \frac{1\overline{)\mathrm{atm}}}{101325\overline{)\mathrm{Pa}}}}\mathbf{\right)}}$

Many gases are shipped in high-pressure containers. Consider a steel tank whose volume is 55.0 gallons and which contains O_{2} gas at a pressure of 16,500 kPa at 24 ^{o}C.

At what temperature would the pressure in the tank equal 155.0 atm?

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