Step 1

**$\mathbf{PV}\mathbf{}\mathbf{=}\mathbf{}\mathbf{nRT}\phantom{\rule{0ex}{0ex}}\mathbf{PV}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{m}}{\mathbf{MM}}\mathbf{RT}\phantom{\rule{0ex}{0ex}}\mathbf{m}\mathbf{=}\frac{\mathbf{PV}\mathbf{\times}\mathbf{MM}}{\mathbf{RT}}\phantom{\rule{0ex}{0ex}}\mathbf{m}\mathbf{}\mathbf{=}\mathbf{}\frac{(16500\mathrm{kPa}\times {\displaystyle \frac{{10}^{3}\mathrm{Pa}}{1\mathrm{kPa}}}\times {\displaystyle \frac{1\mathrm{atm}}{101325\mathrm{Pa}}})(55.0\mathrm{gal}\times {\displaystyle \frac{3.785L}{1\mathrm{gal}}})(32.00{\displaystyle \frac{g}{\mathrm{mol}}})}{(0.08206{\displaystyle \frac{\mathrm{atm}\xb7L}{\mathrm{mol}\xb7K}})(24\xb0C+273.15)\mathbf{K}}$**

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.

What volume would the gas occupy at STP?

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