A gas mixture contains oxygen and argon at partial pressures of 0.60 atm and
425 mmhg. If nitrogen gas added to the sample increases the total pressure to 1250 torr, what is the partial pressure in Torr of the Nitrogen added?
Steps for solving this problem are greatly aprreciated!!!
PO2 = 0.6 atm x (760 mm/1 atm) = ?mm
PAr = 425 mm
Ptotal = 1250 mm
PN2 = 1250 - PO2 - PAr
nitrogen in torr
To solve this problem, we need to apply Dalton's Law of Partial Pressures, which states that the total pressure of a gas mixture is equal to the sum of the partial pressures of each individual gas component.
1. Convert the given partial pressures to the same unit:
- The partial pressure of oxygen is given as 0.60 atm.
- The partial pressure of argon is given as 425 mmHg.
Since mmHg and torr are equivalent units, we can convert the partial pressure of argon to torr:
425 mmHg x (1 torr / 1 mmHg) = 425 torr.
2. Calculate the initial total pressure:
To find the initial total pressure, we sum the partial pressures of oxygen and argon:
Initial total pressure = Partial pressure of oxygen + Partial pressure of argon
Initial total pressure = 0.60 atm + 425 torr = 0.60 atm + 425 torr.
3. Calculate the partial pressure of nitrogen:
To find the partial pressure of nitrogen, we subtract the initial total pressure from the final total pressure:
Partial pressure of nitrogen = Final total pressure - Initial total pressure
Partial pressure of nitrogen = 1250 torr - (0.60 atm + 425 torr).
4. Convert the partial pressure of nitrogen to the desired unit:
Since the given partial pressures are in torr, we can express the partial pressure of nitrogen in torr.
Combining these steps, the solution is as follows:
Partial pressure of nitrogen = (1250 torr) - (0.60 atm + 425 torr) = 1250 torr - (0.60 atm + 425 torr).
Simplifying the expression further will give the final partial pressure of nitrogen in torr.
To solve this problem, we can use Dalton's law of partial pressures, which states that the total pressure exerted by a mixture of gases is the sum of the partial pressures of each individual gas.
1. Start by converting all the pressures to the same units. In this case, let's convert everything to Torr, since the final partial pressure is required in Torr.
- 0.60 atm is equal to (0.60 atm)*(760 Torr/atm) = 456 Torr
- 425 mmHg is already in Torr since 1 mmHg = 1 Torr
- 1250 torr remains the same
2. Apply Dalton's law of partial pressures to find the partial pressure of nitrogen gas:
- Total pressure = partial pressure of oxygen + partial pressure of argon + partial pressure of nitrogen
- 1250 Torr = 456 Torr + 425 Torr + partial pressure of nitrogen
- Solve for the partial pressure of nitrogen:
partial pressure of nitrogen = 1250 Torr - 456 Torr - 425 Torr
partial pressure of nitrogen = 369 Torr
Therefore, the partial pressure of the nitrogen gas added is 369 Torr.