Problem 3 - Chemistry in 30 days

Question

The density of a gas at \(27^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) is \(d\). Pressure remaining constant, at which of the following temperatures will its density become \(0.75 d ?\) (a) \(20^{\circ} \mathrm{C}\) (b) \(30^{\circ} \mathrm{C}\) (c) \(400 \mathrm{~K}\) (d) \(300 \mathrm{~K}\)

Step-by-Step Solution

Answer

The density becomes \(0.75d\) at \(225 \, \text{K}\), none of the given options.

Step 1: Understanding the Ideal Gas Law

The ideal gas law is given by the formula \( PV = nRT \). Here, \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is the temperature in Kelvin.

Step 2: Relating Density to Ideal Gas Law

Density \( d \) can be expressed as \( \frac{mass}{volume} \). From the ideal gas law, \( n = \frac{mass}{molar \ mass} \) and \( V = \frac{nRT}{P} \). Therefore, \( d = \frac{mass}{V} = \frac{P \times molar \ mass}{RT} \).

Step 3: Applying the Given Conditions

We know the density at \( T_1 = 300 \, \text{K} (27^{\circ} \text{C}) \). So, \( d = \frac{P \times molar \ mass}{R \times 300} \). For the new density \( 0.75d \), we use \( T_2 \) as the temperature. So, \( 0.75d = \frac{P \times molar \ mass}{R \times T_2} \).

Step 4: Setting Up the Equation

Equate the expressions for density: \[ \frac{P \times molar \ mass}{R \times 300} \times 0.75 = \frac{P \times molar \ mass}{R \times T_2} \]. Simplify the expression to obtain \( 0.75 \times 300 = T_2 \).

Step 5: Solving for the Temperature

Calculate \( T_2 \): \( T_2 = 0.75 \times 300 = 225 \, \text{K} \).

Step 6: Convert Temperature (Optional)

There is no need to convert to Celsius since \( 225 \, \text{K} \) is directly compared with the options, where the only Kelvin option is \( 300 \, \text{K} \). The answer is computed in Kelvin.

Key Concepts

Gas Density

Gas density is like the oomph of a gas. It tells us how much gas mass is packed into a certain volume. When thinking of gases, we often picture light, airy substances. But their density can change depending on conditions like pressure and temperature. A gas's density is calculated using the formula:

• Density, \( d = \frac{\text{mass}}{\text{volume}} \).

For gases, this is a bit different from solids. That's because gases can be compressed or expanded easily. In the context of the ideal gas law, the density \( d \) can be rewritten using the equation \( d = \frac{P \times \text{molar mass}}{RT} \), with \( P \) being the pressure, \( R \) the ideal gas constant, and \( T \) the temperature in Kelvin. This offers a direct link between density and the ideal gas law parameters. Notice how as temperature increases (and pressure remains constant), the density of a gas decreases. That's because the gas expands, taking up more space.

Temperature Conversion

Converting temperature between Celsius and Kelvin is essential in chemistry, especially when using the ideal gas law. Here, calculations require temperature to be in Kelvin.The formula for conversion is simple:

• To change Celsius to Kelvin: \( T(K) = T(°C) + 273.15 \).

In the given problem, the original temperature is \(27^{\circ} \text{C}\) which converts to \(300 \text{K}\) since it’s mathematically easier to round settings to whole numbers in chemistry.Temperature in Kelvin ensures that calculations are consistent, as Kelvin doesn’t dip below zero. Therefore, any scientific equation works even when temperatures are at their coldest.

Molar Mass Calculation

Molar mass is a crucial property in chemistry that helps determine the connection between mass and moles of a substance. It tells us how much one mole of a substance weighs in grams and is typically expressed in units of g/mol.To find molar mass, you simply add up the atomic masses of all the atoms in a molecule. For example, if calculating the molar mass of hydrogen gas (\(\text{H}_2\)):

• The atomic mass of hydrogen = 1 g/mol.
• So, \(\text{H}_2 = 2 \times 1 = 2 \text{ g/mol} \).

In an ideal gas scenario using the equation \( PV = nRT \), the molar mass plays a part in finding the number of moles \( n \) when combined with density calculations, showcasing different aspects of gas behavior.

Pressure and Volume Relationship

Understanding how pressure and volume relate is crucial for mastering gas behaviors. According to Boyle’s Law, which is a part of the ideal gas laws, the pressure and volume of a given amount of gas are inversely proportional when the temperature remains constant. This means:

• As one increases, the other decreases.
• The equation describing this relationship is \( P_1V_1 = P_2V_2 \).

In many practical cases with gases, temperature can also influence volume and pressure. When thinking about how changes in pressure affect the volume (or vice versa), it helps to consider how flexible a gas can be compared to solids and liquids.For instance, in a sealed bag of chips on an airplane, the bag inflates as pressure decreases outside, illustrating gas laws at work. These principles are part of the overarching ideal gas law formula, applying both for fundamental understanding and practical applications.