Chapter 6: Problem 67
A gas mixture contains oxygen and argon at partial pressures of \(0.60 \mathrm{~atm}\) and \(425 \mathrm{mmHg}\). 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?
Short Answer
Expert verified
369 torr
Step by step solution
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Gas Laws
Gas laws are crucial for understanding how gases behave under various conditions. These laws relate variables such as pressure, volume, and temperature. One of the most fundamental gas laws is Boyle's Law, which states that the pressure of a gas is inversely proportional to its volume when temperature remains constant:
\[ P_1 V_1 = P_2 V_2 \]
Another important law is Charles's Law, which shows that the volume of a gas is directly proportional to its temperature (in Kelvin) when pressure is constant:
\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]
For our particular problem, Dalton's Law of Partial Pressures is most relevant. According to Dalton's Law:
\[ P_{total} = P_1 + P_2 + P_3 + \text{...} \]
This law helps us understand that the total pressure of a gas mixture is the sum of the partial pressures of each component gas.
\[ P_1 V_1 = P_2 V_2 \]
Another important law is Charles's Law, which shows that the volume of a gas is directly proportional to its temperature (in Kelvin) when pressure is constant:
\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]
For our particular problem, Dalton's Law of Partial Pressures is most relevant. According to Dalton's Law:
\[ P_{total} = P_1 + P_2 + P_3 + \text{...} \]
This law helps us understand that the total pressure of a gas mixture is the sum of the partial pressures of each component gas.
Partial Pressure
Partial pressure refers to the pressure that a single gas component in a mixture would exert if it were alone in the container. It is an essential concept when dealing with gas mixtures. You can think of partial pressure as the contribution of each individual gas to the total pressure. According to Dalton’s Law mentioned above, each gas in a mixture adds to the total pressure based on its partial pressure.
For example, in our exercise, the mixture contains oxygen, argon, and nitrogen. The partial pressures of oxygen and argon can be converted, summed up, and finally, the partial pressure of nitrogen can be figured out by subtracting the initial total pressure from the new total pressure.
For example, in our exercise, the mixture contains oxygen, argon, and nitrogen. The partial pressures of oxygen and argon can be converted, summed up, and finally, the partial pressure of nitrogen can be figured out by subtracting the initial total pressure from the new total pressure.
Pressure Unit Conversion
Pressure unit conversion is a necessary skill to solve many gas law problems. Pressure can be measured in various units such as atmospheres (atm), millimeters of mercury (mmHg), and torr. Sometimes, you may need to convert these units to get a consistent measurement. For example:
- 1 atmosphere (atm) is equal to 760 millimeters of mercury (mmHg).
To solve our problem, it was crucial to convert the oxygen's partial pressure from atm to torr (1 torr = 1 mmHg):\[0.60 \text{ atm} \times 760 \text{ torr/atm} = 456 \text{ torr} \]This conversion ensures all pressures are in the same unit and can be accurately added together.
- 1 atmosphere (atm) is equal to 760 millimeters of mercury (mmHg).
To solve our problem, it was crucial to convert the oxygen's partial pressure from atm to torr (1 torr = 1 mmHg):\[0.60 \text{ atm} \times 760 \text{ torr/atm} = 456 \text{ torr} \]This conversion ensures all pressures are in the same unit and can be accurately added together.
Total Pressure
Total pressure is the sum of the partial pressures of all gases present in a mixture. Understanding this concept helps solve problems involving gas mixtures. In our exercise, we calculated the total pressure by adding the partial pressures of oxygen and argon.
Starting with the given pressures:
\[ 0.60 \text{ atm} \times 760 (\text{conversion to torr}) = 456 \text{ torr (O}_2\text{)} \]\[ 425 \text{ torr (Ar)} \] Adding these together gives the initial total pressure:
\[ 456 \text{ torr} + 425 \text{ torr} = 881 \text{ torr} \]To find the partial pressure of the nitrogen added, subtract this initial total pressure from the final total pressure:
\[ 1250 \text{ torr (final total pressure)} - 881 \text{ torr (initial total pressure)} = 369 \text{ torr (N}_2\text{)} \]This shows how each gas contributes to the overall pressure in the container.
Starting with the given pressures:
\[ 0.60 \text{ atm} \times 760 (\text{conversion to torr}) = 456 \text{ torr (O}_2\text{)} \]\[ 425 \text{ torr (Ar)} \] Adding these together gives the initial total pressure:
\[ 456 \text{ torr} + 425 \text{ torr} = 881 \text{ torr} \]To find the partial pressure of the nitrogen added, subtract this initial total pressure from the final total pressure:
\[ 1250 \text{ torr (final total pressure)} - 881 \text{ torr (initial total pressure)} = 369 \text{ torr (N}_2\text{)} \]This shows how each gas contributes to the overall pressure in the container.
