Dimensional Analysis Worksheet 2 Answer Key

Dimensional Analysis Worksheet 2 Answer Key: A thorough look for Students

Introduction

When tackling physics or chemistry problems, dimensional analysis is the essential tool that turns abstract numbers into meaningful quantities. Worksheet 2, a common assignment in high‑school and introductory college courses, often presents a series of conversion and calculation problems that test students’ ability to apply the method systematically. This article provides the complete answer key for Worksheet 2, explains the reasoning behind each solution, and offers practical tips so you can master dimensional analysis on your own.

1. Overview of Worksheet 2 Problems

The worksheet typically contains five sections:

1. Unit Conversions – Convert given quantities between SI and customary units.
2. Dimensional Consistency Checks – Verify whether equations are dimensionally balanced.
3. Rate Law Calculations – Determine reaction rates using stoichiometric coefficients.
4. Density and Mass Calculations – Apply the density formula to find missing variables.
5. Dimensional Analysis for Physical Laws – Use unit analysis to derive useful relationships.

Below, each problem is listed with its answer and a step‑by‑step explanation.

2. Answer Key with Detailed Explanations

2.1 Unit Conversions

Explanation:
Each conversion uses a conversion factor—a ratio equal to 1 expressed in different units. Multiply the given value by the factor to cancel the original unit and leave the desired one.

2.2 Dimensional Consistency Checks

Explanation:
Dimensional analysis requires every term in an equation to have the same set of base dimensions (M, L, T, I, Θ, N, J). Problem 4 is inconsistent because force (N) divided by area (m²) yields pressure (Pa), not a torque (N·m).

2.3 Rate Law Calculations

1. Problem: For the reaction ( 2A + B \rightarrow 3C ), the rate law is ( r = k[A]^2[B] ). If ([A] = 0.1,\text{M}) and ([B] = 0.05,\text{M}), find ( r ) when ( k = 2.5,\text{M}^{-2}\text{s}^{-1} ).
   Answer: ( r = 2.5 \times (0.1)^2 \times 0.05 = 0.00125,\text{M s}^{-1} ) Small thing, real impact. Practical, not theoretical..

2. Problem: A second‑order reaction has rate ( r = k[A][B] ). If ([A] = 0.4,\text{M}), ([B] = 0.2,\text{M}), and ( k = 1.2,\text{M}^{-1}\text{s}^{-1} ), what is the rate?
   Answer: ( r = 1.2 \times 0.4 \times 0.2 = 0.096,\text{M s}^{-1} ) It's one of those things that adds up..

3. Problem: A first‑order reaction has ( k = 0.03,\text{s}^{-1} ). If the initial concentration of A is ( 0.5,\text{M} ), find the concentration after 10 s.
   Answer: ( [A] = 0.5 e^{-0.03 \times 10} = 0.5 e^{-0.3} \approx 0.374,\text{M} ).

4. Problem: For the reaction ( A \rightarrow B ), the rate law is ( r = k[A] ). If ( k = 0.1,\text{s}^{-1} ) and ([A] = 0.2,\text{M}), what is ( r )?
   Answer: ( r = 0.1 \times 0.2 = 0.02,\text{M s}^{-1} ) Turns out it matters..

5. Problem: A reaction follows ( r = k[A]^2 ). If ([A] = 0.3,\text{M}) and ( r = 0.009,\text{M s}^{-1} ), find ( k ).
   Answer: ( k = \frac{r}{[A]^2} = \frac{0.009}{0.09} = 0.1,\text{M}^{-1}\text{s}^{-1} ).

Explanation:
Always match the concentration units to the order of the reaction in the rate law. For first‑order reactions, the rate constant has units of s⁻¹; for second‑order, s⁻¹ M⁻¹; for third‑order, s⁻¹ M⁻², and so on Worth keeping that in mind. Less friction, more output..

2.4 Density and Mass Calculations

1. Problem: A metal cube has a density of ( 7.8,\text{g/cm}^3 ) and a side length of ( 5,\text{cm} ). What is its mass?
   Answer: Volume ( V = 5^3 = 125,\text{cm}^3 ). Mass ( m = \rho V = 7.8 \times 125 = 975,\text{g} ) That's the whole idea..

2. Problem: A liquid has a density of ( 1.05,\text{g/mL} ). How many liters of this liquid weigh ( 10,\text{kg} )?
   Answer: Convert mass to grams: (10,\text{kg} = 10,000,\text{g}). Volume ( V = \frac{m}{\rho} = \frac{10,000}{1.05} \approx 9523.81,\text{mL} = 9.524,\text{L} ).

3. Problem: A rectangular prism has dimensions ( 10,\text{cm} \times 4,\text{cm} \times 3,\text{cm} ) and a density of ( 2.5,\text{g/cm}^3 ). Find its mass.
   Answer: Volume ( V = 10 \times 4 \times 3 = 120,\text{cm}^3 ). Mass ( m = 2.5 \times 120 = 300,\text{g} ) Simple as that..

4. Problem: A spherical balloon has a radius of ( 15,\text{cm} ) and a density of ( 0.0008,\text{kg/m}^3 ). What is its mass?
   Answer: Convert radius to meters: ( 0.15,\text{m} ). Volume ( V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (0.15)^3 \approx 0.01414,\text{m}^3 ). Mass ( m = \rho V = 0.0008 \times 0.01414 \approx 0.0000113,\text{kg} = 0.0113,\text{g} ).

5. Problem: A substance has a mass of ( 250,\text{g} ) and a density of ( 2.5,\text{g/cm}^3 ). What is its volume?
   Answer: ( V = \frac{m}{\rho} = \frac{250}{2.5} = 100,\text{cm}^3 ).

Explanation:
Remember that density is mass per unit volume, so rearranging the formula ( \rho = \frac{m}{V} ) yields ( m = \rho V ) or ( V = \frac{m}{\rho} ). Always keep units consistent; convert centimeters to meters or liters to cubic meters when necessary.

2.5 Dimensional Analysis for Physical Laws

1. Problem: Verify that the kinetic energy formula ( KE = \frac{1}{2}mv^2 ) has dimensions of energy.
   Answer:
   ( [m] = M )
   ( [v] = [L][T]^{-1} )
   ( [v^2] = [L]^2[T]^{-2} )
   ( [KE] = M \times L^2 T^{-2} = [M L^2 T^{-2}] ), which is the dimension of energy (joule).

2. Problem: Show that the pressure formula ( P = \frac{F}{A} ) yields dimensions of force per area.
   Answer:
   ( [F] = M L T^{-2} )
   ( [A] = L^2 )
   ( [P] = \frac{M L T^{-2}}{L^2} = M L^{-1} T^{-2} ), the dimension of pressure (pascal) The details matter here. Surprisingly effective..

3. Problem: Derive the time constant for an RC circuit ( \tau = RC ) using dimensional analysis.
   Answer:
   Resistance ( R ) has units ( \Omega = V/A = (M L^2 T^{-3} I^{-1}) / (M L^2 T^{-3} I^{-1}) = \text{ohm} ). Capacitance ( C ) has units ( F = C/V = I T / V = (I T) / (M L^2 T^{-3} I^{-1}) = M^{-1} L^{-2} T^4 I^2 ). Multiplying ( R \times C ) yields ( T ), confirming that ( \tau ) is a time constant.

4. Problem: Confirm that the ideal gas law ( PV = nRT ) is dimensionally consistent.
   Answer:
   ( [P] = M L^{-1} T^{-2} )
   ( [V] = L^3 )
   ( [n] = \text{mol} ) (dimensionless for stoichiometric purposes)
   ( [R] = M L^2 T^{-2} \Theta^{-1} )
   ( [T] = \Theta )
   Left side: ( M L^2 T^{-2} ).
   Right side: ( n \times R \times T = \text{mol} \times M L^2 T^{-2} \Theta^{-1} \times \Theta = M L^2 T^{-2} ).
   Both sides match.

5. Problem: Determine the units of the diffusion coefficient ( D ) in Fick’s law ( J = -D \frac{dC}{dx} ).
   Answer:
   ( [J] = \text{mol} , \text{m}^{-2} , \text{s}^{-1} )
   ( [\frac{dC}{dx}] = \frac{\text{mol} , \text{m}^{-3}}{\text{m}} = \text{mol} , \text{m}^{-4} )
   Thus, ( [D] = \frac{J}{dC/dx} = \frac{\text{mol} , \text{m}^{-2} , \text{s}^{-1}}{\text{mol} , \text{m}^{-4}} = \text{m}^2 , \text{s}^{-1} ) Nothing fancy..

Explanation:
Dimensional analysis not only checks consistency but can also help derive units of derived quantities without experimental data.

3. Common Mistakes and How to Avoid Them

• Unit Mismatch: Always convert all quantities to the same system (SI or customary) before performing calculations.
• Dropping Conversion Factors: When using a factor like 1 mi = 1609.34 m, write it as ( \frac{1609.34,\text{m}}{1,\text{mi}} ) to ensure the miles cancel.
• Ignoring Exponents: For squared or cubed terms, apply the exponent to the entire conversion factor.
• Forgetting Dimensional Homogeneity: Before solving, write the dimensions of each side of an equation; they must match.
• Misinterpreting Rate Constants: Remember that the units of the rate constant depend on the reaction order; check the reaction order before assigning units.

4. Practical Tips for Mastering Dimensional Analysis

1. Create a Personal Conversion Sheet – Keep the most common conversion factors handy.
2. Practice with “What‑If” Scenarios – Vary the numbers and repeat calculations to build muscle memory.
3. Use Dimensional Checks Early – Before plugging numbers, verify the equation’s dimensional consistency.
4. Label Units Clearly – When writing intermediate steps, attach units to each variable.
5. Teach Someone Else – Explaining the process reinforces your own understanding.

5. Frequently Asked Questions (FAQ)

6. Conclusion

Mastering the dimensional analysis worksheet not only guarantees correct answers but also deepens your understanding of how physical quantities interrelate. Practice regularly, keep a tidy conversion table, and always double‑check dimensions before finalizing your answer. By applying the step‑by‑step methods outlined here—unit conversions, dimensional consistency checks, rate law calculations, density problems, and physical law verification—you’ll build a solid skill set that serves well in academics and beyond. Happy calculating!