Question

Gas Laws The volume \(V\) of an ideal gas varies directly with the temperature \(T\) and inversely with the pressure \(P\). Write an equation relating \(V, T,\) and \(P\) using \(\underline{k}\) as the constant of proportionality. If a cylinder contains oxygen at a temperature of \(300 \mathrm{~K}\) and a pressure of 15 atmospheres in a volume of 100 liters, what is the constant of proportionality \(k ?\) If a piston is lowered into the cylinder, decreasing the volume occupied by the gas to 80 liters and raising the temperature to \(310 \mathrm{~K},\) what is the gas pressure?

Short answer

The constant of proportionality \( k \) is 5 and the new pressure \( P' \) is 19.375 atmospheres.

Step-by-step solution

Determine the relationship between V, T, and P

The volume of an ideal gas varies directly with temperature and inversely with pressure. This can be expressed as: \[ V = k \frac{T}{P} \]where \( k \) is the constant of proportionality.

Solve for the constant of proportionality k

Use the initial conditions to solve for \( k \). Given: \[ T = 300 \, \mathrm{K} \] \[ P = 15 \, \text{atm} \] \[ V = 100 \, \text{liters} \] Substitute these values into the equation: \[ 100 = k \frac{300}{15} \] Simplify this equation to find \( k \): \[ 100 = k \times 20 \] \[ k = \frac{100}{20} = 5 \]

Set up the new conditions equation

With the constant \( k \) known, use the new conditions to set up the equation. Given: \[ V' = 80 \text{ liters} \] \[ T' = 310 \, \mathrm{K} \] We need to find the new pressure \( P' \). From the relationship, we know: \[ V' = k \frac{T'}{P'} \] Substitute \( k = 5 \) into the equation: \[ 80 = 5 \frac{310}{P'} \]

Solve for the new pressure

Isolate \( P' \) in the equation: \[ 80 = 5 \frac{310}{P'} \] \[ 80P' = 1550 \] \[ P' = \frac{1550}{80} \] \[ P' = 19.375 \text{ atm} \]

Key concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry and physics used to describe the behavior of ideal gases. It combines several gas relationships into one equation: \[ PV = nRT \]. Here's what each variable represents:

• \(P\) is the pressure of the gas.
• \(V\) is the volume the gas occupies.
• \(n\) is the number of moles of the gas.
• \(R\) is the ideal gas constant, typically \(8.314 \, \text{J/(mol·K)}\) or equivalent depending on units used.
• \(T\) is the temperature in Kelvin.

This equation provides a comprehensive way to relate the physical properties of gases, taking into account pressure, volume, temperature, and the quantity of the gas. In the exercise, the relationship given is a specific form of the Ideal Gas Law focusing on direct and inverse variations.

Direct and Inverse Variation

Understanding direct and inverse variation helps explain how changes in temperature and pressure affect the volume of a gas.

• **Direct Variation:** This means one variable increases with the increase of another variable. In this case, the volume \(V\) of the gas increases directly with the temperature \(T\).
• **Inverse Variation:** This means one variable increases as the other decreases. Here, the volume \(V\) decreases as the pressure \(P\) increases.

Therefore, we can relate these concepts through the equation:
\[ V = k \frac{T}{P} \], where \( k \) is the constant of proportionality. This equation beautifully combines direct and inverse variations into a simple relationship.

Proportionality Constant

The proportionality constant, \(k\), represents a fixed number that relates the variables in the equation. It's unique to each gas sample based on its specific conditions.
To find \(k\) in the exercise, we use the given initial conditions: \( V = 100 \, \text{liters} \), \( T = 300 \, \text{K} \), and \( P = 15 \, \text{atm} \). Substitute these values into the equation to solve for \(k\):
\[ 100 = k \frac{300}{15} \]
Simplifying, we get:
\[ 100 = 20k \]
Hence:
\[ k = 5 \]
This proportionality constant can now be used to find other unknowns under different conditions.

Solving for Variables

Once the proportionality constant, \(k\), is known, we can solve for other variables using the same fundamental relationship.
In Step 3 of the exercise, the problem asks us to find the gas pressure \( P' \) after changes in volume and temperature, with:

• New volume: \( V' = 80 \, \text{liters} \)
• New temperature: \( T' = 310 \, \text{K} \)
• Known \( k = 5 \)

Using the equation:
\[ V' = k \frac{T'}{P'} \]
Substitute the known values:
\[ 80 = 5 \frac{310}{P'} \]
Solving for \( P' \) involves isolating it:
\[ 80P' = 1550 \]
\[ P' = \frac{1550}{80} \]
\[ P' = 19.375 \, \text{atm} \]
We determine that the new pressure is approximately \( 19.375 \, \text{atm} \). By manipulating the equation properly, you can easily find any unknown variable if the others are known.