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?

Short Answer

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

Step by step solution

01

- 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.
02

- Calculating the number of moles

Continuing from the formula:\[ n = \frac{1.3}{24.63} \]\[ n \approx 0.05278 \, \text{moles} \]
03

- 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 \( n \approx 0.05278 \, \text{moles} \):\[ M = \frac{15}{0.05278} \]\[ M \approx 284.2 \, \text{g/mol} \]
04

- 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:\[ m_{molecule} = \frac{M}{N_A} \]\[ m_{molecule} = \frac{284.2}{6.022 \times 10^{23}} \]\[ m_{molecule} \approx 4.72 \times 10^{-22} \, \text{g} \]
05

- Converting grams to atomic units

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

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

A heat engine uses blackbody radiation as its operating substance. The equation of state for blackbody radiation is \(P=1 / 3 a T^{4}\) and the internal energy is \(U=a V T^{4}\), where \(a=7.566 \times 10^{-16}\) \(\mathrm{J} /\left(\mathrm{m}^{3} \mathrm{~K}^{4}\right)\) is Stefan's constant, \(P\) is pressure, \(T\) is temperature, and \(V\) is volume. The engine cycle consists of three steps. Process \(1 \rightarrow 2\) is an expansion at constant pressure \(P_{1}=P_{2} .\) Process \(2 \rightarrow 3\) is a decrease in pressure from \(P_{2}\) to \(P_{3}\) at constant volume \(V_{2}=V_{3}\). Process \(3 \rightarrow 1\) is an adiabatic contraction from volume \(V_{3}\) to \(V_{1}\). Assume that \(P_{1}=3.375 P_{3}, T_{1}=2000 \mathrm{~K}\), and \(V_{1}=10^{-3} \mathrm{~m}^{3}\). (a) Express \(V_{2}\) in terms of \(V_{1}\) and \(T_{1}=T_{2}\) in terms of \(T_{3}\) (b) Compute the work done during each part of the cycle. (c) Compute the heat absorbed during each part of the cycle. (d) What is the efficiency of this heat engine (get a number)? (e) What is the efficiency of a Carnot engine operating between the highest and lowest temperatures.

A monatomic fluid in equilibrium is contained in a large insulated box of volume \(V\). The fluid is divided (conceptually) into \(m\) cells, each of which has an average number of particles \(N_{0}\), where \(N_{0}\) is large (neglect coupling between cells). Compute the variance in fluctuations of internal energy per particle \(u=U / N,\left\langle\left(\Delta u_{i}\right)^{2}\right\rangle\), in the ith cell. (Hint: Use temperature \(T\) and volume per particle \(v=V / N\) as independent variables.)

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A solution of particles A and B has a Gibbs free energy $$ \begin{aligned} G\left(P, T, \mathrm{n}_{\mathrm{A}}, \mathrm{n}_{\mathrm{B}}\right)=& \mathrm{n}_{\mathrm{A}} g_{\mathrm{A}}(P, T)+\mathrm{n}_{\mathrm{B}} g_{\mathrm{B}}(P, T)+\frac{1}{2} \lambda_{\mathrm{AA}} \frac{\mathrm{n}_{\mathrm{A}}^{2}}{\mathrm{n}}+\frac{1}{2} \lambda_{\mathrm{BB}} \frac{\mathrm{n}_{\mathrm{B}}^{2}}{\mathrm{n}} \\ &+\lambda_{\mathrm{AB}} \frac{\mathrm{n}_{\mathrm{A}} \mathrm{n}_{\mathrm{B}}}{\mathrm{n}}+\mathrm{n}_{\mathrm{A}} R T \ln x_{\mathrm{A}}+\mathrm{n}_{\mathrm{B}} R T \ln x_{\mathrm{B}} \end{aligned} $$ Initially, the solution has \(\mathrm{n}_{\mathrm{A}}\) moles of A and \(\mathrm{n}_{\mathrm{B}}\) moles of B. (a) If an amount \(\Delta \mathrm{n}_{\mathrm{B} \text {, of } \mathrm{B} \text { is added }}\) keeping the pressure and temperature fixed, what is the change in the chemical potential of A? (b) For the case \(\lambda_{\mathrm{AA}}=\lambda_{\mathrm{BB}}=\lambda_{\mathrm{AB} \text {, }}\) does the chemical potential of A increase or decrease?

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