Question

At a certain temperature for which \(\mathrm{RT}=25 \mathrm{~L}\) atm. \(\mathrm{mol}^{-1}\), the density of a gas, in \(\mathrm{g} \mathrm{L}^{-1}\), is \(\mathrm{d}=2.00 \mathrm{P}+\) \(0.020 \mathrm{P}^{2}\), where \(\mathrm{P}\) is the pressure in atmosphere. The molecular weight of the gas in \(\mathrm{g} \mathrm{mol}-1\) is (a) 60 (b) 75 (c) 50 (d) 35

Short answer

The molecular weight of the gas is 50 g/mol.

Step-by-step solution

Understand the Problem

We need to find the molecular weight of the gas using the given density equation: \(d = 2.00P + 0.020P^2\). We are informed that \(RT = 25 \; \text{L atm mol}^{-1}\).

Express the Ideal Gas Law for Density

The ideal gas law \(PV = nRT\) can be modified for density: \(P = \frac{dRT}{M}\), where \(M\) is the molar mass (molecular weight) of the gas. We need to manipulate this to find \(M\).

Substitute Given for Density into the Ideal Gas Expression

Replace \(d\) in the ideal gas law formula with \(d = 2.00P + 0.020P^2\) to get \(P = \frac{(2.00P + 0.020P^2) \cdot RT}{M}\).

Simplify and Solve for Molecular Weight \(M\)

The equation becomes \(P = \frac{25 \times (2.00P + 0.020P^2)}{M}\). Solving for \(M\), multiply both sides by \(M\) and divide by \(P\): \(M = 25 \times (2.00 + 0.020P) = 50 + 0.5P\). The pressure \(P\) cancels, as it must hold for all \(P\), simplifying to \(M = 50\).

Match with Given Options

The molecular weight calculated is 50 g/mol, so the correct option is (c) 50.

Key concepts

Ideal Gas Law

The ideal gas law is a fundamental equation for understanding the behavior of gases under various conditions. It is expressed as \(PV = nRT\), where:

• \( P \) is the pressure of the gas in atmospheres (atm)
• \( V \) is the volume of the gas in liters (L)
• \( n \) is the number of moles of gas
• \( R \) is the ideal gas constant, \( 0.0821 \, \text{L atm mol}^{-1} \text{K}^{-1} \)
• \( T \) is the temperature in Kelvin (K)

This equation is crucial because it establishes a relationship among pressure, volume, temperature, and amount of gas.
To utilize this for finding the density of a gas, we can manipulate the equation. Density \( d \) is defined as mass divided by volume, typically in \( \text{g/L} \). Substituting \( n = \frac{m}{M} \) (where \( m \) is the mass and \( M \) is the molar mass), the ideal gas law can be rewritten in terms of density. This gives us the formula for density: \( d = \frac{PM}{RT} \). Thus, understanding the ideal gas law allows us to manipulate these parameters to derive useful quantities like density and molar mass.

Density of Gases

Density is a key property of gases and helps us determine their mass relative to volume. It is especially useful for identifying gases and calculating their molecular weights.
In the context of the ideal gas law, the density \( d \) of a gas is expressed as \( d = \frac{PM}{RT} \). This formula shows that:

• Density is directly proportional to the pressure \( P \) and the molar mass \( M \)
• Density is inversely proportional to the product \( RT \)

For example, if we have a gas where the density equation is \( d = 2.00P + 0.020P^2 \), we can substitute this into the ideal gas expression to find the relationship between the density and pressure.
The challenge comes from manipulating these mathematical expressions to solve for one variable, like molecular weight, which is why understanding how density changes with pressure and temperature (through \( RT \)) is critically important for deeper gas analysis.

Molar Mass Determination

Determining the molar mass (or molecular weight) of a gas is an essential application of chemistry that involves using the density and the ideal gas law. The molar mass \( M \) is calculated through the rearrangement of the formula we derived for density: \( d = \frac{PM}{RT} \).
This allows us to isolate \( M \) and express it as: \[ M = \frac{dRT}{P} \].
In our step-by-step solution, we showed that substituting \( d = 2.00P + 0.020P^2 \) into this equation lets us derive an expression for \( M \) as \( M = 50 + 0.5P \). Here, the pressure \( P \) effect cancella at all points, clarifying that the molar mass is consistent and independent of \( P \).
Once you have the formula in hand, choosing the right results from given options becomes straightforward. Thus, determining molar mass involves both conceptual understanding and technical manipulation of equations—emphasizing practical skills in math and chemistry.