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A 2.50-L container is filled with \(175 \mathrm{~g}\) argon. a. If the pressure is \(10.0\) atm, what is the temperature? b. If the temperature is \(225 \mathrm{~K}\), what is the pressure?

Short Answer

Expert verified
a. The temperature of the argon gas is \(T = \frac{PV}{nR} = \frac{10.0 \text{ atm} \cdot 2.50 \text{ L}}{\frac{175}{39.95} \text{ mol} \cdot 0.0821 \frac{\text{L.atm}}{\text{mol.K}}} = 298.15 \text{ K}\). b. The pressure of the argon gas is \(P = \frac{nRT}{V} = \frac{\frac{175}{39.95} \text{ mol} \cdot 0.0821 \frac{\text{L.atm}}{\text{mol.K}} \cdot 225 \text{ K}}{2.50 \text{ L}} = 4.95 \text{ atm}\).

Step by step solution

01

Find the moles of Argon gas

Given mass of Argon = 175 g To find the number of moles, we will use the molar mass of Argon, which is 39.95 g/mol. Moles (n) = mass / molar_mass n = \( \frac{175}{39.95} \) Step 2: Calculate temperature
02

Calculate the temperature

We use the Ideal gas law: PV = nRT Given, P = 10.0 atm, V = 2.50 L, R = 0.0821 \( \frac{\text{L.atm}}{\text{mol.K}} \), solving for T, T = \( \frac{PV}{nR} \) Step 3: Calculate Pressure
03

Calculate the pressure

Again, we use the Ideal gas law: PV = nRT Given, T = 225 K, V = 2.50 L, R = 0.0821 \( \frac{\text{L.atm}}{\text{mol.K}} \), solving for P, P = \( \frac{nRT}{V} \)

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Key Concepts

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

Argon Gas
Argon is a noble gas, which means it is inert and does not react readily with other substances. This characteristic makes argon useful in various applications such as filling light bulbs and in welding to provide an inert atmosphere. When dealing with gases like argon in a container, it is critical to understand how it behaves under different conditions of pressure, temperature, and volume.
  • Argon is a monoatomic gas, consisting of single atoms, which makes the molar mass calculation straightforward.
  • Its molar mass is 39.95 g/mol, which is a crucial number when you need to convert between grams and moles in chemical calculations.
  • This conversion is essential when using equations like the ideal gas law, as we often calculate with moles rather than mass.
Moles Calculation
Calculating the number of moles of a gas is an important step in using gas laws, such as the ideal gas law. The number of moles refers to the amount of substance present, and it is crucial for understanding the behavior of gases in chemical reactions and different conditions.
To calculate moles, you use the formula:
- Moles ( n ) = mass / molar mass
For example, if you have 175 grams of argon gas, and you know the molar mass of argon is 39.95 g/mol, you can find the moles by calculating:
\[n = \frac{175 \text{ g}}{39.95 \text{ g/mol}}\approx 4.38 \text{ moles of argon}\].
  • After finding the moles, you can then input this value into various equations and calculations required for the study of gas dynamics.
  • This step is foundational for leveraging the ideal gas law in predicting how argon or any other gas behaves under different conditions.
Pressure and Temperature Relationship
The ideal gas law, expressed as \( PV = nRT \), is key to understanding the relationship between pressure, volume, and temperature for gases. This law applies particularly well to ideal gases, a model which most real gases approximate under many conditions.
  • When applying the ideal gas law, if you know any three of the variables (pressure \(P\), volume \(V\), number of moles \(n\), gas constant \(R\), temperature \(T\)), you can calculate the fourth.
  • The constant \(R\) varies based on the units used, but a common value is 0.0821 \( \frac{L\cdot atm}{mol\cdot K} \) for pressure in atm and volume in liters.
  • Temperature must always be in Kelvin when using this equation, as it avoids negative numbers and provides a direct measure proportional to the gas's energetic temperature.
For instance, to find temperature \(T\) if pressure is 10 atm and volume is 2.5 L, with the number of moles previously calculated:
\[T = \frac{PV}{nR} = \frac{10 \times 2.5}{4.38 \times 0.0821} \approx 69 \text{ K} \].
Similarly, to find the pressure \(P\) when temperature is 225 K, it follows:
\[P = \frac{nRT}{V} = \frac{4.38 \times 0.0821 \times 225}{2.5} \approx 32 \text{ atm} \].
Understanding these calculations gives insight into how a gas like argon responds to changes in its environment.

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Most popular questions from this chapter

Use the following information to identify element \(\mathrm{A}\) and compound \(\mathrm{B}\), then answer questions a and \(\mathrm{b}\). An empty glass container has a mass of \(658.572 \mathrm{~g} .\) It has a mass of \(659.452 \mathrm{~g}\) after it has been filled with nitrogen gas at a pressure of 790 . torr and a temperature of \(15^{\circ} \mathrm{C}\). When the container is evacuated and refilled with a certain element (A) at a pressure of 745 torr and a temperature of \(26^{\circ} \mathrm{C}\), it has a mass of \(660.59 \mathrm{~g}\) Compound \(\mathrm{B}\), a gaseous organic compound that consists of \(85.6 \%\) carbon and \(14.4 \%\) hydrogen by mass, is placed in a stainless steel vessel \((10.68 \mathrm{~L})\) with excess oxygen gas. The vessel is placed in a constant-temperature bath at \(22^{\circ} \mathrm{C}\). The pressure in the vessel is \(11.98 \mathrm{~atm}\). In the bottom of the vessel is a container that is packed with Ascarite and a desiccant. Ascarite is asbestos impregnated with sodium hydroxide; it quantitatively absorbs carbon dioxide: $$ 2 \mathrm{NaOH}(s)+\mathrm{CO}_{2}(g) \longrightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{H}_{2} \mathrm{O}(l) $$ The desiccant is anhydrous magnesium perchlorate, which quantitatively absorbs the water produced by the combustion reaction as well as the water produced by the above reaction. Neither the Ascarite nor the desiccant reacts with compound \(\mathrm{B}\) or oxygen. The total mass of the container with the Ascarite and desiccant is \(765.3 \mathrm{~g}\) The combustion reaction of compound \(\mathrm{B}\) is initiated by a spark. The pressure immediately rises, then begins to decrease, and finally reaches a steady value of \(6.02 \mathrm{~atm} .\) The stainless steel vessel is carefully opened, and the mass of the container inside the vessel is found to be \(846.7 \mathrm{~g}\). \(\mathrm{A}\) and \(\mathrm{B}\) react quantitatively in a \(1: 1\) mole ratio to form one mole of the single product, gas \(\mathrm{C}\). a. How many grams of \(\mathrm{C}\) will be produced if \(10.0 \mathrm{~L} \mathrm{~A}\) and \(8.60 \mathrm{~L}\) \(\mathrm{B}\) (each at STP) are reacted by opening a stopcock connecting the two samples? b. What will be the total pressure in the system?

A steel cylinder contains \(5.00 \mathrm{~mol}\) graphite (pure carbon) and \(5.00 \mathrm{~mol} \mathrm{O}_{2} .\) The mixture is ignited and all the graphite reacts. Combustion produces a mixture of \(\mathrm{CO}\) gas and \(\mathrm{CO}_{2}\) gas. After the cylinder has cooled to its original temperature, it is found that the pressure of the cylinder has increased by \(17.0 \% .\) Calculate the mole fractions of \(\mathrm{CO}, \mathrm{CO}_{2}\), and \(\mathrm{O}_{2}\) in the final gaseous mixture.

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The steel reaction vessel of a bomb calorimeter, which has a volume of \(75.0 \mathrm{~mL}\), is charged with oxygen gas to a pressure of 145 atm at \(22^{\circ} \mathrm{C}\). Calculate the moles of oxygen in the reaction vessel.

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