Question

How much work (in \(\mathrm { J }\) ) is required to expand the volume of a pump from 0.0\(\mathrm { L }\) to 2.5\(\mathrm { L }\) against an external pressure of 1.1\(\mathrm { atm }\) ?

Short answer

The work required is -275 J.

Step-by-step solution

Convert Volume

First, convert the volume from liters to cubic meters. 1 liter is equivalent to 0.001 cubic meters.

Convert Pressure

Next, convert pressure from atmospheres to Pascals. 1 atmosphere is equivalent to 101,325 Pascals.

Calculate Work Done

Now, use the equation for work done against constant external pressure: Work = - Pressure * Change in Volume. The work will be negative as the system does work against the surroundings.

Key concepts

Work-Energy Principle in Chemistry

The work-energy principle in chemistry is fundamentally about the transfer of energy between systems and their surroundings due to mechanical work. In the context of gases and thermodynamics, work is typically done when a gas expands or contracts.

When a gas expands, it must push against external pressure and thus does work on its surroundings. Conversely, if it's being compressed, work is done on the gas. The work done is related to the force applied over a distance, which in terms of gases, translates to the pressure applied over a change in volume.

This principle is key in understanding processes like chemical reactions in closed systems, where pressure-volume work can indicate whether energy is absorbed or released during the reaction. It is also a crucial concept for devices that involve gas expansions and compressions, such as heat engines or refrigerators.

Volume Expansion Work

When we talk about volume expansion work in chemistry, we are looking at a scenario where a gas expands within a container. The work done by the gas during expansion depends on the external pressure and the change in volume.

To calculate the work done, we typically use the equation: \( W = - P_{\text{ext}} \times \text{Δ}V \), where \(P_{\text{ext}} \) is the external pressure and \( \text{Δ}V \) is the change in volume. The sign convention here is that work is taken as negative when a system expands against external pressure because energy is being used to do work on the surroundings.

Pressure-Volume Work

Pressure-volume work, or P-V work, is a type of work that occurs when the volume of a system changes in the presence of an external pressure. This concept is central to thermodynamics and is described by the first law, which combines the concepts of internal energy, heat, and work.

In a situation where a gas expands against a constant pressure, we calculate work by the formula: \( W = - P_{\text{ext}} \times \text{Δ}V \). The negative sign indicates that the system is losing energy as it does work on its surroundings. It is important to remember that in real-world scenarios, external pressures can fluctuate, which makes the calculation of work more complex and often requires integration if the pressure changes with volume.

Unit Conversion in Chemistry Calculations

Unit conversion is a pivotal step in chemistry calculations, including when calculating work in thermodynamics. As seen in the example exercise, working with standard SI units is crucial. Volume often needs conversion from liters (\( L \) ) to cubic meters (\( m^3 \) ), and pressure from atmospheres (\( atm \) ) to Pascals (\( Pa \) ).

To convert volume, we use the fact that 1 \( L \) is equal to 0.001 \( m^3 \) . For pressure, we use that 1 \( atm \) is approximately equal to 101,325 \( Pa \) . Getting these conversions correct is essential because using inconsistent or incorrect units will lead to erroneous calculations of work, which can have significant implications in experimental and industrial processes.