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If 0.500 mol of hydrogen gas occupies 5.00 L at \(25^{\circ} \mathrm{C}\), what is the pressure in atmospheres?

Short Answer

Expert verified
The pressure is approximately 2.45 atm.

Step by step solution

01

Identify what you know

You have 0.500 moles of hydrogen gas, a volume of 5.00 L, and a temperature of 25°C. Convert the temperature to Kelvin, since gas laws use absolute temperature scales: \[ T = 25 + 273.15 = 298.15 \, \text{K} \].
02

Use the Ideal Gas Law Formula

The ideal gas law equation is \( PV = nRT \), where:- \( P \) is the pressure in atm,- \( V \) is the volume in liters,- \( n \) is the number of moles,- \( R \) is the ideal gas constant \(0.0821 \, \text{L} \, \text{atm} \, \text{mol}^{-1} \, \text{K}^{-1} \)- \( T \) is the temperature in Kelvin (298.15 K in this case).
03

Rearrange the Formula to Solve for Pressure

We want to solve for \( P \), so rearrange the formula: \[ P = \frac{nRT}{V} \].
04

Substitute the Known Values into the Formula

Substitute the known values into the equation:\[ P = \frac{0.500 \times 0.0821 \times 298.15}{5.00} \].
05

Calculate the Pressure

Perform the calculation:\[ P = \frac{0.500 \times 0.0821 \times 298.15}{5.00} \approx 2.45 \, \text{atm} \].

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Key Concepts

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

Gas Pressure Calculations
Understanding how to calculate gas pressure is an essential part of working with gases in chemistry. The pressure of a gas is a force exerted by gas particles colliding with the walls of their container. To calculate it using the Ideal Gas Law, you follow a straightforward formula: - **Formula**: \( P = \frac{nRT}{V} \) - \( P \): Pressure in atmospheres (atm) - \( n \): Number of moles - \( R \): Ideal gas constant (0.0821 L atm mol\(^{-1}\) K\(^{-1}\)) - \( T \): Temperature in Kelvin - \( V \): Volume in litersIn our example, we calculated the pressure of 0.500 mol of hydrogen gas in a 5.00 L container at 25°C. By using the Ideal Gas Law, we rearranged the formula to solve for pressure: \[P = \frac{0.500 \times 0.0821 \times 298.15}{5.00}\]After performing the multiplication and division, the calculated pressure is approximately 2.45 atm. For students, practicing these calculations can significantly improve understanding of how gases behave under different conditions.
Mole Concept in Chemistry
The mole concept is key to understanding chemistry. It relates to the counting of atoms, molecules, or other chemical entities in terms of moles, a standard unit of measurement.- **1 mole** equates to **Avogadro's number**, which is approximately \(6.022 \times 10^{23}\) entities, allowing you to convert between the mass of a substance and the number of particles it contains.In the context of gases, **moles** help relate the amount of gas to other properties like volume, pressure, and temperature. In our example, 0.500 moles indicates that half of \(6.022 \times 10^{23}\) molecules of hydrogen gas occupy the space involved.Through the Ideal Gas Law equation, you can understand how changing this amount (moles) impacts pressure or volume. A good grasp of the mole concept allows you to tackle many problems in chemistry with confidence.
Temperature Conversion in Gas Laws
When dealing with gas laws, temperature is always considered in Kelvin. This absolute temperature scale avoids negative numbers and ensures consistency in calculations. Conversion from Celsius to Kelvin is straightforward by adding 273.15:\[T (K) = T (°C) + 273.15\]In our example, 25°C was converted to 298.15 K. Using Kelvin ensures the energy corresponding to temperature remains positive, crucial for accurate results in gas law calculations. Understanding why Kelvin is used is essential. Gas laws derive from kinetic theory, where temperature is a measure of average kinetic energy. This translates correctly into calculations only if temperatures are in an absolute form, like Kelvin.Always remember: triple-check your conversions to avoid errors in your calculations. Mistakes in temperature conversion are common but easily preventable with practice.

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