Question

If \(20.0 \mathrm{~g}\) of \(\mathrm{N}_{2}\) gas has a volume of \(4.00 \mathrm{~L}\) and a pressure of 6.0 atm, what is its temperature?

Short answer

The temperature is approximately 409 K.

Step-by-step solution

Identify the Problem Type and Known Variables

This problem is related to the gas laws. We are given the mass of nitrogen gas, volume, pressure, and need to find the temperature using the ideal gas law. Known variables are:- Mass of \( \text{N}_2 \): 20.0 g- Volume (V): 4.00 L- Pressure (P): 6.0 atmWe need to convert the given mass to moles to use the ideal gas law.

Convert Mass to Moles

To convert the mass of \( \text{N}_2 \) to moles, use the molar mass of nitrogen gas, which is 28.02 g/mol:\[ n = \frac{\text{mass}}{\text{molar mass}} = \frac{20.0 \text{ g}}{28.02 \text{ g/mol}} \approx 0.714 \text{ moles} \]

Apply the Ideal Gas Law

The ideal gas law is given by \( PV = nRT \), where:- \( P \) is the pressure (6.0 atm)- \( V \) is the volume (4.00 L)- \( n \) is the number of moles (0.714)- \( R \) is the ideal gas constant (0.0821 L atm/mol K)- \( T \) is the temperature in KelvinRearrange the equation to solve for \( T \):\[ T = \frac{PV}{nR} \]

Calculate the Temperature

Substitute the known values into the rearranged ideal gas law:\[ T = \frac{(6.0 \text{ atm})(4.00 \text{ L})}{(0.714 \text{ moles})(0.0821 \text{ L atm/mol K})} \]\[ T \approx \frac{24.0}{0.0586} \approx 409.04 \text{ K} \]

Report the Temperature

The temperature of the \( \text{N}_2 \) gas is approximately 409 K. Since we calculate in Kelvin, there's no need to convert to another scale unless specifically required.

Key concepts

Gas Laws

The gas laws are a set of rules describing the behavior of gases under various conditions of pressure, volume, and temperature. A key law that often comes into play is the Ideal Gas Law, which is pivotal in chemistry and physics. This law is expressed as \( PV = nRT \), where:

• \(P\) is the pressure of the gas,
• \(V\) is the volume occupied by the gas,
• \(n\) is the amount of substance (in moles),
• \(R\) is the ideal gas constant, approximately \(0.0821 \text{ L atm/mol K}\),
• \(T\) is the temperature in Kelvin.

This equation allows us to solve for one variable if the others are known. For instance, in the given problem, we use it to determine the temperature of nitrogen gas when we have its pressure, volume, and number of moles. Understanding and implementing this equation allows for predictability in how gases will react under different circumstances, making it a fundamental concept in science.

Moles Conversion

Converting mass to moles is a crucial step when utilizing the Ideal Gas Law. This conversion is necessary because the law uses the number of moles, not mass, to calculate other conditions like temperature or pressure. Here's how you convert mass to moles:

• Identify the substance and find its molar mass. For \( \text{N}_2 \), the molar mass is \(28.02 \text{ g/mol}\).
• Divide the mass of the sample by its molar mass. So, \(n = \frac{20.0 \text{ g}}{28.02 \text{ g/mol}} \approx 0.714 \text{ moles}\).

This process transforms the units from grams to moles, which are more practical for calculations involving the Ideal Gas Law. By understanding how to convert between mass and moles, you gain a valuable tool in solving a broad range of chemical problems.

Temperature Calculation

Calculating temperature using the Ideal Gas Law is straightforward once you have all the necessary information. After converting mass to moles and identifying the pressure and volume, you rearrange the Ideal Gas Law to solve for temperature \((T)\). Here's how it works:

• Start with the equation \(PV = nRT\).
• Rearrange to find \(T\): \(T = \frac{PV}{nR}\).
• Substitute known values: \(P = 6.0 \text{ atm}\), \(V = 4.00 \text{ L}\), \(n = 0.714 \text{ moles}\), \(R = 0.0821 \text{ L atm/mol K}\).

The calculation becomes \(T = \frac{(6.0)(4.00)}{(0.714)(0.0821)}\), resulting in approximately \(409.04 \text{ K}\). Temperatures calculated in Kelvin will not require conversion unless specified differently by the problem. The calculation provides insights into the energy state of the gas, as temperature is directly related to the kinetic energy of particles.