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How much work (in J) is involved in a chemical reaction if the volume decreases from \(33.6 \mathrm{~L}\) to \(11.2 \mathrm{~L}\) against a constant pressure of \(90.5 \mathrm{kPa}\) ?

Short Answer

Expert verified
The work involved in the chemical reaction is \(2026.8 \mathrm{~J}\).

Step by step solution

01

Convert pressure to Pa

The pressure is given in kPa, but the SI unit for pressure is Pascal (Pa). To convert kPa to Pa, we can multiply the given pressure with 1000 since 1 kPa = 1000 Pa. So, the pressure in Pa is: 90.5 kPa × 1000 = 90500 Pa
02

Calculate the change in volume

We are given the initial volume (33.6 L) and the final volume (11.2 L). The change in volume ΔV is the difference between final and initial volumes: ΔV = V_final - V_initial ΔV = 11.2 L - 33.6 L ΔV = -22.4 L
03

Convert volume to m³

The SI unit for volume is cubic meters (m³), so we need to convert the volume ΔV from liters to m³. We can use the conversion factor that 1 L = 0.001 m³: ΔV_m³ = ΔV(L) × 0.001 ΔV_m³ = -22.4 L × 0.001 ΔV_m³ = -0.0224 m³
04

Calculate work done

Now, we can use the formula W = -PΔV to calculate the work done: W = - (90500 Pa) × (-0.0224 m³) W = 2026.8 J So, the work involved in the chemical reaction is 2026.8 J.

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Key Concepts

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

Work Done in Chemistry
In chemistry, the concept of work done is pivotal when studying reactions involving gases. When a gas expands or contracts, it exerts a force on its surroundings, and this is where the concept of work comes in. Work done (W) quantifies the energy transferred when a force moves an object. In chemical terms, work done refers to the amount of energy exchanged due to volume changes during reactions. For gases, this involves pressure and volume changes which are often described using the formula:
\[ W = - P \Delta V \]
This equation signifies that the work done (W) is equal to the negative of the external pressure (P) multiplied by the change in volume (\Delta V). The negative sign in the formula represents that work done by the system on its surroundings is considered negative. Conversely, if the system is compressed, work is done on the system, making it positive. Understanding this formula is crucial as it highlights the interplay between energy, work, and chemical changes that can often occur in reaction chambers or controlled environments.
  • Work is a transfer of energy.
  • In chemistry, it often involves gases and their volumes.
  • The sign of work determines the direction of energy transfer.
Pressure-Volume Work
Pressure-volume work is a specific type of work done by or on the system when a chemical reaction occurs involving gases. It deals with changes in volume under constant pressure. Imagine you have a piston that compresses or expands within a cylinder as a gas inside reacts. This mechanical movement due to pressure differences reflects pressure-volume work.
This concept is essential in thermodynamics, as it allows us to understand how energy, in the form of work, helps drive chemical processes.
  • Pressure remains constant during the reaction.
  • A change in volume (\(\Delta V\)) indicates work done.
  • This work is energy transferred due to volume changes.
By using the formula \(W = - P \Delta V\), we can calculate the amount of work done during chemical processes involving gases. Changes in volume can either be due to expansion, where the system does work on the surroundings, or compression, where the surroundings do work on the system. It serves as an excellent model for processes like the operation of internal combustion engines or even biological lungs during breathing.
Unit Conversion in Chemistry
Unit conversion is a crucial step in chemistry, especially when dealing with calculations involving work done, pressure, and volume. The primary goal of conversion is to ensure that all units match SI standards for accurate calculation. In the study exercise, converting pressure from kPa to Pa and volume from L to m³ was necessary for using the pressure-volume work formula.
Here's how to tackle unit conversion effectively:
  • Always convert pressure to Pascals (Pa) – the SI unit of pressure – because 1 kPa equals 1000 Pa.
  • Convert volumes to cubic meters (m³) when calculating work, as 1 L is equivalent to 0.001 m³.
  • Consistent units throughout the calculation help prevent errors and yield accurate results.
Thus, converting units is more than just a step in calculations; it's a fundamental practice that ensures results are both accurate and meaningful in the context of scientific analysis and understanding. Proper unit conversion is a mark of precision and care in chemistry, reinforcing the basic principle that consistent units enable effective problem-solving.

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Most popular questions from this chapter

You may have noticed that when you compress the air in a bicycle pump, the body of the pump gets warmer. (a) Assuming the pump and the air in it comprise the system, what is the sign of \(w\) when you compress the air? (b) What is the sign of \(q\) for this process? (c) Based on your answers to parts (a) and (b), can you determine the sign of \(\Delta E\) for compressing the air in the pump? If not, what would you expect for the sign of \(\Delta E\) ? What is your reasoning? [Section 5.2]

Consider a system consisting of the following apparatus, in which gas is confined in one flask and there is a vacuum in the other flask. The flasks are separated by a valve. Assume that the flasks are perfectly insulated and will not allow the flow of heat into or out of the flasks to the surroundings. When the valve is opened, gas flows from the filled flask to the evacuated one. (a) Is work performed during the expansion of the gas? (b) Why or why not? (c) Can you determine the value of \(\Delta E\) for the process?

Consider the following hypothetical reactions: $$ \begin{array}{l} \mathrm{A} \longrightarrow \mathrm{B} \quad \Delta H_{I}=+60 \mathrm{~kJ} \\ \mathrm{~B} \longrightarrow \mathrm{C} \quad \Delta H_{I I}=-90 \mathrm{~kJ} \end{array} $$ (a) Use Hess's law to calculate the enthalpy change for the reaction \(\mathrm{A} \longrightarrow \mathrm{C}\). (b) Construct an enthalpy diagram for substances A, B, and C, and show how Hess's law applies.

The hydrocarbons cyclohexane \(\left(\mathrm{C}_{6} \mathrm{H}_{12}(l), \Delta H_{f}^{\circ}=-156\right.\) \(\mathrm{kJ} / \mathrm{mol}\) ) and 1-hexene \(\left(\mathrm{C}_{6} \mathrm{H}_{12}(l), \Delta H_{f}^{\circ}=-74 \mathrm{~kJ} / \mathrm{mol}\right)\) have the same empirical formula. (a) Calculate the standard enthalpy change for the transformation of cyclohexane to 1-hexene. (b) Which has greater enthalpy, cyclohexane or 1-hexene? (c) Without doing a further calculation and knowing the answer to (b), do you expect cyclohexane or 1-hexene to have the larger combustion enthalpy?

A house is designed to have passive solar energy features. Brickwork incorporated into the interior of the house acts as a heat absorber. Each brick weighs approximately \(1.8 \mathrm{~kg}\). The specific heat of the brick is \(0.85 \mathrm{~J} / \mathrm{g}-\mathrm{K} .\) How many bricks must be incorporated into the interior of the house to provide the same total heat capacity as \(1.7 \times 10^{3}\) gal of water?

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