Question

The density of oxygen gas at \(25^{\circ} \mathrm{C}\) is \(1.458 \mathrm{mg} /\) litre at one atmosphere. \(\Lambda t\) what pressure will oxygen have the density twice the value? (1) \(0.5 \mathrm{~atm} 25^{\circ} \mathrm{C}\) (2) \(2 \mathrm{~atm}, 25^{\circ} \mathrm{C}\) (3) \(4 \mathrm{~atm}, 25^{\circ} \mathrm{C}\) (4) None

Short answer

2 atm, 25°C.

Step-by-step solution

- Understand the relationship

The density of a gas is directly proportional to its pressure when temperature is held constant. This is derived from the ideal gas law: \( PV = nRT \), where \(d = \frac{m}{V}\) and \(n = \frac{m}{M}\). Thus, \( d = \frac{PM}{RT} \).

- Calculate the new density

We are given that the new density is twice the original density. So, the new density is \(2 \times 1.458 \mathrm{mg/L} = 2.916 \mathrm{mg/L}\).

- Set up the proportion

Using the relationship, \(d \propto P\), we can set up the proportion: \( \frac{d_2}{d_1} = \frac{P_2}{P_1} \). Substitute the known values: \( \frac{2.916 \mathrm{mg/L}}{1.458 \mathrm{mg/L}} = \frac{P_2}{1 \mathrm{atm}} \).

- Solve for the new pressure

Solve the proportion: \( 2 = \frac{P_2}{1 \mathrm{atm}} \). Therefore, \( P_2 = 2 \mathrm{atm} \).

- Choose the correct answer

The answer that matches our calculated pressure is (2) 2 atm, 25°C.

Key concepts

Density of Gases

Density is a measure of how much mass is contained in a given volume. It is commonly represented by the symbol \(d\), given by the formula: \(d = \frac{m}{V}\), where \(m\) is the mass and \(V\) is the volume.
For gases, density changes based on pressure and temperature. When temperature is held constant, the density of a gas is directly proportional to the pressure exerted on it.
This means, as pressure increases, density also increases, and vice versa. This idea comes from the Ideal Gas Law, which we will discuss next.

Pressure and Temperature Relationship

The Ideal Gas Law is a fundamental equation in chemistry and physics, represented as \( PV = nRT \).
Here, \(P\) is pressure, \(V\) is volume, \(n\) is the amount of gas in moles, \(R\) is the gas constant, and \(T\) is temperature in Kelvin.
According to this law:

• Pressure is directly proportional to temperature when the volume and amount of gas are constant.
• As temperature increases, the kinetic energy of gas molecules increases, causing more frequent and forceful collisions, which increases pressure.

However, in our exercise, temperature remains constant at \(25^{\text{°}} \text{C}\), so we observe how pressure affects density.

Direct Proportionality in Gases

In gases, direct proportionality means that if one variable doubles, the other doubles as well. From the Ideal Gas Law, we can isolate density \(d\) as: \[ d = \frac{PM}{RT} \] where \(M\) is the molar mass.

• When pressure \(P\) increases, density \(d\) also increases directly, assuming temperature \(T\) and molar mass \(M\) are constant.
• If gas density doubles, the pressure must also double under constant temperature.

In the given exercise, the original density was \(1.458\) mg/L. Doubling the density leads to \(2.916\) mg/L. Using the relationship \( \frac{d_2}{d_1} = \frac{P_2}{P_1} \), doubling the density means the pressure doubles to \(2 \text{ atm}\).