Problem 30 - A Modern Course in Statistical Physics

Question

A biological molecule of unknown mass can be prepared in pure powdered form. If 15 \(\mathrm{g}\) of this powder is added to a container with \(1 \mathrm{~L}\) of water at \(T=300 \mathrm{~K}\), which is initially at atmospheric pressure, the pressure inside the container increases to \(P=1.3 \mathrm{~atm}\). (a) What is the molecular weight of the biological molecules? (b) What is the mass of each molecule expressed in atomic units?

Step-by-Step Solution

Answer

The molecular weight is approximately 284.2 g/mol. The mass of each molecule is approximately 284.3 atomic mass units (AMU).

Step 1 - Understanding the Ideal Gas Law

Use the Ideal Gas Law equation: \[ PV = nRT \]where \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the gas constant, and \( T \) is the temperature. Here we need to find \( n \), the number of moles of the gas.

Step 3 - Calculating the number of moles

Continuing from the formula:\[ n = \frac{1.3}{24.63} \]\[ n \approx 0.05278 \, \text{moles} \]

Step 4 - Finding the Molecular Weight

Molecular weight (M) is found using the formula:\[ M = \frac{m}{n} \]where \( m \) is the mass and \( n \) is the number of moles.Given \( m = 15 \, \text{g} \) and \( n \approx 0.05278 \, \text{moles} \):\[ M = \frac{15}{0.05278} \]\[ M \approx 284.2 \, \text{g/mol} \]

Step 5 - Converting to Atomic Units

To find the mass of each molecule in atomic mass units, use the Avogadro's number \( N_A = 6.022 \times 10^{23} \, \text{mol}^{-1}\). The mass of each molecule \( m_{molecule} \) is calculated as:\[ m_{molecule} = \frac{M}{N_A} \]\[ m_{molecule} = \frac{284.2}{6.022 \times 10^{23}} \]\[ m_{molecule} \approx 4.72 \times 10^{-22} \, \text{g} \]

Step 6 - Converting grams to atomic units

Since 1 atomic mass unit (AMU) is approximately \( 1.6605 \times 10^{-24} \, \text{g} \):\[ m_{molecule} \approx \frac{4.72 \times 10^{-22}}{1.6605 \times 10^{-24}} \]\[ m_{molecule} \approx 284.3 \, \text{AMU} \]

Key Concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry and physics that describes the behavior of an ideal gas. The equation is expressed as: \[ PV = nRT \] where:

• \( P \) 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 gas constant (usually 0.0821 L·atm/(K·mol)).
• \( T \) is the temperature in Kelvin.

In this problem, you are given the pressure \( P = 1.3 \, \text{atm} \), volume \( V = 1 \, \text{L} \), and temperature \( T = 300 \, \text{K} \). Using the Ideal Gas Law, we can calculate the number of moles \( n \) of the gas. Rearrange the formula to solve for \( n \): \[ n = \frac{PV}{RT} \] Plugging in the values:\[ n = \frac{1.3 \, \text{atm} \times 1 \, \text{L}}{0.0821 \, \text{L·atm/(K·mol)} \times 300 \, \text{K}} \] This will give you the amount of substance in moles, which is crucial for calculating the molecular weight later.

Molecular Weight Calculation

Molecular weight (or molar mass) is the mass of one mole of a substance, and it is expressed in grams per mole \( \text{g/mol} \). We can find it using the mass \( m \) and number of moles \( n \) of the substance. The formula is: \[ M = \frac{m}{n} \] For this exercise, you have already calculated the number of moles \( n \) and are given the mass \( m \) of the biological molecule (15 grams). Inserting the values: \[ M = \frac{15 \, \text{g}}{0.05278 \, \text{moles}} \] \[ M \approx 284.2 \, \text{g/mol} \] This value tells you that one mole of this biological molecule weighs approximately 284.2 grams.

Atomic Mass Units

Atomic mass units (AMUs) are a standard unit of mass that quantifies mass on an atomic or molecular scale. One AMU is defined as precisely \( 1/12 \) of the mass of a carbon-12 atom, which is about \( 1.6605 \times 10^{-24} \, \text{g} \). To convert the mass of a molecule from grams to AMUs, you use Avogadro's number \( N_A = 6.022 \times 10^{23} \, \text{mol}^{-1} \) which tells you the number of molecules in a mole.The mass of each molecule in grams is: \[ m_{molecule} = \frac{M}{N_A} \] Substituting the molecular weight: \[ m_{molecule} = \frac{284.2 \, \text{g/mol}}{6.022 \times 10^{23} \, \text{mol}^{-1}} \] \[ m_{molecule} \approx 4.72 \times 10^{-22} \, \text{g} \] To convert this to AMUs: \[ m_{molecule} \approx \frac{4.72 \times 10^{-22} \, \text{g}}{1.6605 \times 10^{-24} \, \text{g/AMU}} \] \[ m_{molecule} \approx 284.3 \, \text{AMU} \] So each biological molecule weighs about 284.3 AMUs.