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A \(2.50-\mathrm{L}\) flask contains 0.60 \(\mathrm{g} \mathrm{O}_{2}\) at a temperature of \(22^{\circ} \mathrm{C}\) . What is the pressure (in atm) inside the flask?

Short Answer

Expert verified
The pressure inside the flask is approximately \(0.184\,\text{atm}\).

Step by step solution

01

1. Convert the temperature from Celsius to Kelvin

To convert the temperature from Celsius to Kelvin: \(T_K = T_C + 273.15\) In this case, the transformation will be: \(22^{\circ}\mathrm{C} + 273.15\) K. So, \(T_K = 295.15\,\text{K}\).
02

2. Convert the given mass of gas to moles

To find the number of moles, we'll use the molecular weight of oxygen, O2 (32 g/mol): \(n = \frac{\text{mass}}{\text{molecular weight}}\) \(n = \frac{0.60\,\text{g}}{32\,\frac{\text{g}}{\text{mol}}}\) \(n \approx 0.01875\,\text{mol}\)
03

3. Calculate the pressure using the Ideal Gas Law

Now that we have the number of moles of gas (\(n \approx 0.01875\,\text{mol}\)), the temperature in Kelvin (\(T_K = 295.15\,\text{K}\)), and the flask volume (\(V = 2.50\,\text{L}\)), we can calculate the pressure using the Ideal Gas Law formula: \(P=\frac{nRT}{V}\) \(P=\frac{(0.01875\,\text{mol})(0.0821\,\frac{\text{L}\cdot\text{atm}}{\text{K}\cdot\text{mol}})(295.15\,\text{K})}{2.50\,\text{L}}\) \(P \approx 0.184\,\text{atm}\) So, the pressure inside the flask is approximately \(0.184\,\text{atm}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Pressure Calculation
Calculating the pressure of a gas requires understanding the Ideal Gas Law. This law is expressed with the formula \( P = \frac{nRT}{V}\), where \( P \) represents pressure, \( n \) stands for the number of moles, \( R \) is the gas constant (0.0821 \(\frac{L\cdot atm}{K\cdot mol}\)), \( T \) is temperature in Kelvin, and \( V \) is the volume in liters.
  • Plug into the formula all known quantities.
  • Ensure that temperature is in Kelvin and volume is in liters for consistency with \( R \).
  • Calculate \( P \) by substituting the values for \( n \), \( R \), \( T \), and \( V \).
In our exercise, these values are: \( n \approx 0.01875\,\text{mol} \), \( R \approx 0.0821\,\frac{L\cdot atm}{K\cdot mol} \), \( T \approx 295.15\,\text{K} \), and \( V = 2.50\,\text{L} \). Calculating for \( P \) gives approximately \( 0.184\,\text{atm} \). As we can see, following these steps leads us directly to the solution.
Molar Mass
Understanding molar mass is crucial for converting a substance's mass into moles. Molar mass is the mass of one mole of a substance and is usually expressed in grams per mole (\(g/mol\)). For molecular oxygen (\( O_2 \)), the molar mass is \( 32\, g/mol \).
  • To find moles, use the formula: \( n = \frac{\text{mass}}{\text{molar mass}} \).
  • In this exercise, the mass of \( O_2 \) is \( 0.60\, g \).
  • Dividing the mass by \( 32\, g/mol \) gives us \( \approx 0.01875\, mol \).
Accurate mole calculation is essential, as it is directly used in the Ideal Gas Law to find pressure. Recognizing the role of molar mass helps bridge the gap between mass and the Ideal Gas components.
Temperature Conversion
Converting temperature from Celsius to Kelvin is a common need in chemistry, especially when using the Ideal Gas Law. Since the law requires Kelvin, we follow a simple formula: \( T_K = T_C + 273.15 \).
  • Kelvin is an absolute temperature scale based on absolute zero.
  • Celsius is more commonly used in everyday contexts but needs conversion for scientific formulas.
In the given exercise, the temperature of \(22^{\circ} \text{C} \) is converted to Kelvin by simply adding \( 273.15 \), resulting in \( 295.15\,\text{K} \). Accurately converting temperatures ensures correct calculations in pressure estimation using the Ideal Gas Law.
Oxygen Gas
Oxygen gas (\( O_2 \)) is a diatomic molecule carrying significant importance in chemistry. Its properties, such as molar mass and reactivity, often make it a focus of study in gas calculations like those in the Ideal Gas Law.
  • As a diatomic molecule, each consists of two oxygen atoms.
  • Molar mass of \( O_2 \) is \( 32\, g/mol \), arising from two atoms of oxygen, each approximately \( 16\, g/mol \).
In this particular exercise, understanding \( O_2 \) helps us not only calculate the moles from grams but also appreciate its behavior in different conditions, affecting pressure and volume.
Gas Volume
Gas volume relates directly to how much space gas occupies, here measured in liters. When calculating aspects like pressure using the Ideal Gas Law, knowing the gas volume is pivotal.
  • Gas volume largely depends on container size and is usually constant for the exercise.
  • Ideal Gas Law assumes volume does not change with the gas inside.
In our context, the volume is \(2.50\,\text{L}\). When we plug this into the Ideal Gas Law, it helps us derive the pressure value per the model the law provides. Without an accurate measure of volume, the pressure read would undoubtedly be inaccurate.

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Most popular questions from this chapter

Suppose two \(200.0-\mathrm{L}\) tanks are to be filled separately with the gases helium and hydrogen. What mass of each gas is needed to produce a pressure of 2.70 atm in its respective tank at \(24^{\circ} \mathrm{C} ?\)

A certain flexible weather balloon contains helium gas at a volume of 855 L. Initially, the balloon is at sea level where the temperature is \(25^{\circ} \mathrm{C}\) and the barometric pressure is 730 torr. The balloon then rises to an altitude of 6000 ft, where the pressure is 605 torr and the temperature is \(15^{\circ} \mathrm{C}\). What is the change in volume of the balloon as it ascends from sea level to 6000 ft?

What will be the effect on the volume of an ideal gas if the pressure is doubled and the absolute temperature is halved?

The nitrogen content of organic compounds can be determined by the Dumas method. The compound in question is first reacted by passage over hot \(\mathrm{CuO}(\mathrm{s})\): $${\text { Compound }} \frac{\text { Hot }}{\text { \(\mathrm{CuO}(\mathrm{s})\) }} \mathrm{N}_{2}(g)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)$$ The product gas is then passed through a concentrated solution of \(\mathrm{KOH}\) to remove the \(\mathrm{CO}_{2} .\) After passage through the \(\mathrm{KOH}\) solution, the gas contains \(\mathrm{N}_{2}\) and is saturated with water vapor. In a given experiment a 0.253 -g sample of a compound produced 31.8 \(\mathrm{mL} \mathrm{N}_{2}\) saturated with water vapor at \(25^{\circ} \mathrm{C}\) and 726 torr. What is the mass percent of nitrogen in the compound? (The vapor pressure of water at \(25^{\circ} \mathrm{C}\) is 23.8 torr.

You have a helium balloon at 1.00 atm and \(25^{\circ} \mathrm{C} .\) You want to make a hot-air balloon with the same volume and same lift as the helium balloon. Assume air is 79.0\(\%\) nitrogen and 21.0\(\%\) oxygen by volume. The 鈥渓ift鈥 of a balloon is given by the difference between the mass of air displaced by the balloon and the mass of gas inside the balloon. a. Will the temperature in the hot-air balloon have to be higher or lower than \(25^{\circ} \mathrm{C} ?\) Explain. b. Calculate the temperature of the air required for the hot-air balloon to provide the same lift as the helium balloon at 1.00 atm and \(25^{\circ} \mathrm{C}\) . Assume atmospheric conditions are 1.00 atm and \(25^{\circ} \mathrm{C} .\)

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