10. Answer any two of the parts (a), (b) and (c) - Leaving Cert Chemistry - Question 10 - 2022

Real gases behave more like ideal gases under the following conditions:

1. High temperature: At elevated temperatures, the kinetic energy of the gas particles is higher, reducing the effect of intermolecular forces.
2. Low pressure: At low pressures, the particles are farther apart, minimizing the volume occupied by the gas particles themselves.

The noble gas helium (He) is expected to behave most like an ideal gas. This is because:

1. Helium has the smallest atomic size among noble gases, which means it occupies a smaller volume.
2. It has the fewest electrons and therefore the weakest intermolecular interactions (such as van der Waals forces).
3. Helium can be described as light and non-polar, further supporting its behavior as an ideal gas due to minimal attraction between particles.

To calculate the relative molecular mass (M_r) of the gas, we will use the ideal gas equation: $PV = nRT$.

1. First, we rearrange to find n (number of moles):
   $n = \frac{PV}{RT}$.

   Given:

   • P = 99,000 Pa (which is 99 kPa)
   • V = 60.0 L = 0.060 m³
   • R = 8.314 J/(mol·K)
   • T = 22 °C = 295 K

   Substituting values:
   $n = \frac{(99,000)(0.060)}{(8.314)(295)}$
   $n \approx 2.43 \text{ moles}$.

2. Now, we can find the relative molecular mass:
   $M_r = \frac{m}{n},$ where m is the mass of the gas (4.85 g).
   $M_r = \frac{4.85}{2.43} \approx 2.00$.

Hence, the relative molecular mass of the gas is approximately 2.