At standard temperature and pressure (0 °C and 1.00 atm ), 1.00 mol of an ideal gas occupies a volume of 22.4 L. What volume would the same amount of gas occupy at the same pressure and 90 °C?

Express your answer with the appropriate units.

The ideal gas law (PV = nRT) describes the relationship among pressure *P*, volume *V*, temperature *T*, and molar amount *n*. When *n* and *V* are fixed, the equation can be rearranged to take the following form where *k* is a constant:

$\frac{\mathbf{P}}{\mathbf{T}}\mathbf{=}\frac{\mathbf{nR}}{\mathbf{V}}\mathbf{=}\mathbf{k}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{or}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\left(\frac{P}{T}\right)}_{\mathbf{initial}}\mathbf{=}{\left(\frac{P}{T}\right)}_{\mathbf{final}}$

When *n* and *P* are fixed, the expression becomes

$\frac{\mathbf{V}}{\mathbf{T}}\mathbf{=}\frac{\mathbf{nR}}{\mathbf{P}}\mathbf{=}\mathbf{k}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{or}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\left(\frac{V}{T}\right)}_{\mathbf{initial}}\mathbf{=}{\left(\frac{V}{T}\right)}_{\mathbf{final}}$

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