Problem 28 - Chemistry Matter and Change

Question

What is the pressure, in atmospheres, of a 0.108 -mol sample of helium gas at a temperature of \(20.0^{\circ} \mathrm{C}\) if its volume is 0.505 L?

Step-by-Step Solution

Answer

The pressure is approximately 5.15 atm.

Step 1: Identify the Formula

To find the pressure of a gas, we can use the Ideal Gas Law, which is given by the formula \(PV = nRT\). Here, \(P\) is the pressure, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the ideal gas constant, and \(T\) is the temperature in Kelvin.

Step 2: Convert Temperature to Kelvin

The temperature is given in degrees Celsius, so we need to convert it to Kelvin. Use the formula: \[T(K) = T(^{\circ}C) + 273.15\]. Thus, \(T = 20.0 + 273.15 = 293.15\, \text{K}\).

Step 3: Use the Ideal Gas Constant

The ideal gas constant \(R\) can be used in several units. For this problem, since the volume is in Liters and pressure is to be found in atmospheres, use \(R = 0.0821\, \text{L atm/mol K}\).

Step 4: Rearrange the Ideal Gas Law to Find Pressure

Rearrange the formula \(PV = nRT\) equation to solve for pressure \(P\): \[P = \frac{nRT}{V}\].

Step 5: Insert Known Values Into the Equation

Substitute the known values into the rearranged formula:\(n = 0.108\, \text{mol}\), \(R = 0.0821\, \text{L atm/mol K}\), \(T = 293.15\, \text{K}\), and \(V = 0.505\, \text{L}\). The equation becomes: \[P = \frac{0.108 \times 0.0821 \times 293.15}{0.505}\].

Step 6: Calculate the Pressure

Perform the calculation: \(P = \frac{0.108 \times 0.0821 \times 293.15}{0.505} = \frac{2.598}{0.505} = 5.146\, \text{atm}\). The pressure of the gas is approximately \(5.15\, \text{atm}\).

Key Concepts

Pressure Calculation

To determine the pressure of a gas within a container, we employ a fundamental principle of chemistry known as the Ideal Gas Law. This law is articulated with the formula \( PV = nRT \), which correlates the pressure \( P \), volume \( V \), amount in moles \( n \), ideal gas constant \( R \), and temperature \( T \). By rearranging this equation to \( P = \frac{nRT}{V} \), we can easily isolate and solve for the pressure \( P \). This calculation gives valuable insight into how gases behave under different conditions.

Understanding pressure in atmospheres helps in practical applications, as it allows us to consider how gases will respond in various environments. By substituting the known values for \( n \), \( R \), \( T \), and \( V \) into this rearranged equation, we can quickly and efficiently find the pressure exerted by the gas, which in our example is approximately \( 5.15 \, \text{atm} \).

Key points to remember:

• Use the Ideal Gas Law: \( PV = nRT \)
• Rearrange to find pressure: \( P = \frac{nRT}{V} \)
• Ensure units are consistent for accurate results

Temperature Conversion

Temperature is a crucial aspect of gas law calculations, as it directly influences the kinetic energy and behavior of gas particles. Most scientific calculations use Kelvin as the standard unit for temperature, since Kelvin is an absolute scale with no negative numbers, making it logical for scientific use.

To convert a Celsius temperature to Kelvin, simply apply the formula \[T(K) = T(^{\circ}C) + 273.15\]This simple addition adjusts for whether the temperature is above or below the freezing point. In the example given, the temperature of \( 20.0^{\circ} \mathrm{C} \) converts to \( 293.15 \mathrm{K} \). This important step ensures that all calculations are performed using the proper, universal scale.

Remember:

• Convert Celsius to Kelvin by adding \( 273.15 \)
• Kelvin scale is always used in gas law calculations
• This conversion is vital for precise calculations and interpretations

Gas Constant

The ideal gas constant, denoted as \( R \), is a vital component of the Ideal Gas Law equation. It provides the proportionality to link all variables of the gas law equation. Its value depends on the units used for pressure, volume, and temperature. In most chemical applications, \( R \) is given as \( 0.0821 \, \text{L atm/mol K} \), which corresponds to using liters for volume, atmospheres for pressure, and Kelvin for temperature.

The selection of the gas constant helps streamline calculations and provides compatibility with the other units in use. It’s essential to use the right value of \( R \) to ensure accuracy in any calculation involving the Ideal Gas Law equation.

When using the Ideal Gas Law, keep in mind:

• Select appropriate \( R \) value based on units in use
• Commonly, \( R = 0.0821 \, \text{L atm/mol K} \) in chemistry
• Consistency in units ensures the validity of the results