Question

At the same temperature and pressure, which sample contains the samenumber of moles of particles as 1 liter of O2(g)?(1) 1 L Ne(g) (3) 0.5 L SO2(g)(2) 2 L N2(g) (4) 1 L H2O(ℓ)

Answer 1

Answer:

1 L of Ne (g)

Explanation:

According to the ideal gas law, $pV = nRT$. Rearranging this expression for moles, we would get:

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

For two samples, the following equation should then hold:

$\frac{p_1V_1}{RT_1} = \frac{p_2V_2}{RT_2}$

At the same temperature and pressure, this is then rearranged to:

$V_1 = V_2$

Since we're examining the moles of particles rather than moles of molecules, we may use stoichiometry assuming that the first gas has x atoms in a molecule and that the second gas has y atoms in a molecule:

$xV_1 = yV_2$

Assuming that the first gas is oxygen, we get:

$2V_1 = yV_2$

Given a total of 1 liter:

$2 = yV_2$

• For neon, we get $yV_2 = 1\cdot 1 = 1$;
• For sulfur dioxide, we get $yV_2 = 3\cdot 1 = 3$;
• For nitrogen, we get $yV_2 = 2\cdot 2 = 4$;
• Liquid water doesn't obey the ideal gas law, we exclud it.

None of the options gives the same number of moles of atoms. However, thinking about molecules, it is sufficient to state that:

$V_1 = V_2$

This means 1 liter of neon would have the same number of molecules as 1 liter of oxygen.