State Boyle's law - Leaving Cert Chemistry - Question b - 2009

Using the combined gas law:

$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$

Where:

• Initial Pressure, $P_1 = 2 \times 10^5 \;Pa$
• Initial Volume, $V_1 = 12230 \; cm³$
• Initial Temperature, $T_1 = 298 \; K$
• Final Pressure, $P_2 = 1 \times 10^5 \;Pa$
• Final Temperature, $T_2 = 273 \; K$

Rearranging the formula to find $V_2$:

$V_2 = \frac{P_1 \cdot V_1 \cdot T_2}{P_2 \cdot T_1}$

Substituting the values:

$V_2 = \frac{(2 \times 10^5) \cdot 12230 \cdot 273}{(1 \times 10^5) \cdot 298} \approx 22408 \; cm³$

Using the ideal gas law equation, which is:

$PV = nRT$

Where:

• $P = 1 \times 10^5 \; Pa$
• $V = 22408 \; cm³ = 22408 \times 10^{-6} \; m³ = 0.022408 \; m³$
• $R = 8.314 \; J/(mol \cdot K)$
• $T = 273 \; K$

Rearranging to find $n$ (the number of moles):

$n = \frac{PV}{RT}$

Substituting:

$n = \frac{(1 \times 10^5) \cdot (0.022408)}{(8.314) \cdot (273)} \approx 0.005 \; moles$

To find the number of molecules, we can use Avogadro's number:

$N = n \times N_A$

Where:

• $N_A = 6.022 \times 10^{23} \; molecules/mol$
• $n = 0.005 \; moles$

Thus:

$N = 0.005 \times (6.022 \times 10^{23}) \approx 3.011 \times 10^{21} \; molecules$