Answer :
Approximately 0.0305 wine bottles can be purged with the argon in the canister.
Let's go through the calculations step by step:
Given:
- Pressure (\(P\)) = 1.26 atm
- Volume (\(V\)) = 0.75 L (converted from 750.0 ml)
- Temperature (\(T\)) = 283 K
- Ideal Gas Constant (\(R\)) = 0.0821 L·atm/(mol·K)
Now, use the ideal gas law to find the number of moles [tex](\(n\)):\[ n = \frac{PV}{RT} \][/tex]
[tex]\[ n = \frac{(1.26 \, \text{atm}) \times (0.75 \, \text{L})}{(0.0821 \, \text{L·atm/(mol·K)}) \times (283 \, \text{K})} \]\[ n = \frac{0.945}{23.2263} \]\[ n \approx 0.0408 \, \text{mol} \][/tex]
Now, let's assume that the argon canister contains pure argon, and we need to find out how many moles of argon are in a wine bottle. Argon's molar mass is approximately 39.95 g/mol.
[tex]\[ \text{Moles of argon in a wine bottle} = \frac{\text{Molar mass of argon} \times \text{Volume of the bottle}}{\text{Molar volume at STP}} \]\[ \text{Moles of argon in a wine bottle} = \frac{(39.95 \, \text{g/mol}) \times (0.75 \, \text{L})}{22.4 \, \text{L/mol}} \]\[ \text{Moles of argon in a wine bottle} \approx 1.34 \, \text{mol} \][/tex]
Finally, to find the number of wine bottles that can be purged with the argon in the canister:
[tex]\[ \text{Number of bottles} = \frac{\text{Total moles in canister}}{\text{Moles in one bottle}} \]\[ \text{Number of bottles} = \frac{0.0408 \, \text{mol}}{1.34 \, \text{mol/bottle}} \]\[ \text{Number of bottles} \approx 0.0305 \][/tex]
Therefore, approximately 0.0305 wine bottles can be purged with the argon in the canister.
