Two cannisters are connected by a closed tap. Both cannisters contain a mixture of nitrogen and oxygen. The first cannister has a volume of 5.5 L and is at a pressure of 120 kPa. It contains 32% oxygen. The second cannister has a volume of 6.7 L and is at a pressure of 113.8 kPa. It contains 52% oxygen. What is the partial pressure of the oxygen when the tap is opened.
Question
Two cannisters are connected by a closed tap. Both cannisters contain a mixture of nitrogen and oxygen. The first cannister has a volume of 5.5 L and is at a pressure of 120 kPa. It contains 32% oxygen. The second cannister has a volume of 6.7 L and is at a pressure of 113.8 kPa. It contains 52% oxygen. What is the partial pressure of the oxygen when the tap is opened.
Solution
To solve this problem, we need to use the ideal gas law and the concept of partial pressure.
Step 1: Calculate the total moles of gas in each canister using the ideal gas law (PV = nRT). We can assume a constant temperature and use R = 8.3145 L·kPa/(K·mol).
For canister 1: n1 = P1V1 / RT = (120 kPa * 5.5 L) / (8.3145 L·kPa/(K·mol) * T) = n1T
For canister 2: n2 = P2V2 / RT = (113.8 kPa * 6.7 L) / (8.3145 L·kPa/(K·mol) * T) = n2T
Step 2: Calculate the moles of oxygen in each canister.
For canister 1: nO2_1 = 0.32 * n1T
For canister 2: nO2_2 = 0.52 * n2T
Step 3: Calculate the total moles of oxygen when the tap is opened.
nO2_total = nO2_1 + nO2_2
Step 4: Calculate the total volume and total pressure when the tap is opened.
V_total = V1 + V2 = 5.5 L + 6.7 L = 12.2 L
P_total = (n1T + n2T)RT / V_total
Step 5: Calculate the partial pressure of oxygen.
PO2 = (nO2_total / (n1T + n2T)) * P_total
This will give you the partial pressure of the oxygen when the tap is opened.
Note: This solution assumes that the temperature remains constant and that the gases behave ideally, which may not be the case in reality.
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