We’re going to calculate the pressure of each gas first using the **ideal gas equation**:

$\overline{){\mathbf{PV}}{\mathbf{=}}{\mathbf{nRT}}}$

P = pressure, atm

V = volume, L

n = moles, mol

R = gas constant = 0.08206 (L·atm)/(mol·K)

T = temperature, K

*Rearranging the ideal gas equation:*

$\frac{\mathbf{P}\overline{)\mathbf{V}}}{\overline{)\mathbf{V}}}\mathbf{=}\frac{\mathbf{nRT}}{\mathbf{V}}\phantom{\rule{0ex}{0ex}}\overline{){\mathbf{P}}{\mathbf{=}}\frac{\mathbf{nRT}}{\mathbf{V}}}$

•** Calculate the partial pressure of CO _{2}:**

**mass CO _{2} = 350 g**

The molar mass of CO_{2}:

**CO**_{2} **1 **x 12.01 g/mol = 12.01 g/mol** 2** x 16.00 g/mol __= 32.00 g/mol __** Total: ****44.01 g/mol**

$\mathbf{\hspace{0.17em}}\mathbf{moles}\mathbf{}{\mathbf{CO}}_{\mathbf{2}}\mathbf{=}\mathbf{350}\mathbf{.}\mathbf{0}\mathbf{}\overline{)\mathbf{g}\mathbf{}{\mathbf{CO}}_{\mathbf{2}}}\mathbf{\times}\frac{\mathbf{1}\mathbf{}\mathbf{mol}\mathbf{}{\mathbf{CO}}_{\mathbf{2}}}{\mathbf{44}\mathbf{.}\mathbf{01}\mathbf{}\overline{)\mathbf{g}\mathbf{}{\mathbf{O}}_{\mathbf{2}}}}$

**moles CO _{2} = 7.95 mol CO_{2}**

A 36.0–L cylinder of a gas used for calibration of blood gas analyzers in medical laboratories contains 350 g CO_{2}, 805 g O_{2}, and 4,880 g N_{2}. At 25 degrees C, what is the pressure in the cylinder in atmospheres?

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