Question

Complete the following table for an ideal gas.

 PVnTa.7.74×103Pa12.2 mL 25∘C\begin{matrix} \text{ } & \text{P} & \text{V} & \text{n} & \text{T}\\ \hline \text{a.} & \text{$7.74 \times 10^{3} \mathrm{Pa}$} & \text{12.2 mL} & \text{ } & \text{$25^{\circ} \mathrm{C}$}\\ \end{matrix}

Solution

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Answered 1 year ago
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We simply use and rearrange the ideal gas equation to solve for the missing molar amount (nn). Temperatures should strictly be converted to Kelvin before calculations (25 o^oC = 298.2 K). Volume should be in liters and pressure should be in atm (7.74 ×103\times 10^{3} Pa = 0.0764 atm):

PV=nRTn=PVRT=(0.0764 atm)(0.0122 L)(0.08206 L⋅atmmol⋅K)(298.2 K)=3.81×10−5 mol\begin{align*} PV &= nRT\\ n &= \dfrac{PV}{RT}\\ &= \mathrm{\dfrac{(0.0764~atm)(0.0122~L)}{(0.08206~\frac{L \cdot atm}{mol \cdot K})(298.2~K)}}\\ &= \boxed{\mathrm{3.81 \times 10^{-5} ~mol}}\\ \end{align*}

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