9.6 Ideal Gas Law
Notes
The ideal gas law relates pressure, volume, moles and temperature for one set of conditions.
- $PV = nRT$, with $R = 0.0821$ L·atm/(mol·K).
- Units must match $R$: $P$ in atm, $V$ in liters, $n$ in moles, $T$ in kelvins.
- Example: $\displaystyle n = \frac{PV}{RT} = \frac{(2.00)(10.0)}{(0.0821)(300)} = 0.812$ mol.
- To find molar mass: find $n$ from $PV = nRT$, then $\displaystyle M = \frac{\text{mass}}{n}$.
Practice
What is the pressure of 0.500 mol of gas in a 5.00 L container at 273 K?
$\displaystyle P = \frac{nRT}{V} = \frac{(0.500)(0.0821)(273)}{5.00} = 2.24$ atm.
What volume does 1.00 mol of gas occupy at 1.00 atm and 298 K?
$\displaystyle V = \frac{(1.00)(0.0821)(298)}{1.00} = 24.5$ L. 22.4 L is for 273 K, not 298 K.
A 1.60 g sample of gas fills 1.12 L at 273 K and 1.00 atm. What is its molar mass?
$\displaystyle n = \frac{(1.00)(1.12)}{(0.0821)(273)} = 0.0500$ mol. $\displaystyle M = \frac{1.60}{0.0500} = 32.0$ g/mol.
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