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Answer :
The volume occupied by helium gas at STP is V2 = 0.31L. At the same temperature and pressure the volume of ozone is ~8.13 L.
To answer question 1 regarding the volume of helium gas at STP:
We can use the combined gas law which is P1V1/T1 = P2V2/T2, where P is pressure, V is volume, and T is temperature (in Kelvin). First, convert all units to the correct SI units and STP conditions. Standard temperature and pressure (STP) are defined as 0°C (273 K) and 1 atm (101.3 kPa). Here's the conversion:
The initial volume (V1) = 350 cm³ or 0.350 L
The initial pressure (P1) = 99.3 kPa
The initial temperature (T1) = 22.0°C = 295.15 K (after converting from °C to K)
The final pressure (P2) = 101.3 kPa (at STP)
The final temperature (T2) = 273.15 K (at STP)
Insert these values into the combined gas law and solve for V2 (the final volume):
(99.3 kPa)(0.350 L) / (295.15 K) = (101.3 kPa)(V2) / (273.15 K)
V2 = 0.31L.
Solving for V2 gives us the volume occupied by helium gas at STP.
To address question 2 regarding the volume of ozone:
Ozone (O3) is formed from oxygen (O2) with a 2:3 ratio. This means that for every three molecules of oxygen, two molecules of ozone are formed. Since the amount of substance is conserved, the number of moles will change according to this ratio.
0.5 moles of O2 will form (0.5 moles * 2/3) = ~0.333 moles of O3. Under the same conditions of temperature and pressure, the volume occupied by a gas is directly proportional to the number of moles of the gas (according to Avogadro's law).
Therefore, if 0.5 moles of O2 occupy 12.2 L, then 0.333 moles of O3 will occupy:
(12.2 L * (0.333 moles / 0.5 moles)) = ~8.13 L.
The volume of ozone at the same temperature and pressure after the conversion will be ~8.13 L.
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