Question

How much work (in J) is involved in a chemical reaction if the volume decreases from 5.00 to 1.26 L against a constant pressure of 0.857 atm?

Short answer

The work involved in the chemical reaction is approximately 324.8 J.

Step-by-step solution

Write down the given values and find the volume change

Given volume before the reaction, $V_1=5.00L$; and volume after the reaction, $V_2=1.26L$. First, let's calculate the change in volume. $$\mathrm{\Delta }V=V_2-V_1$$

Calculate the change in volume

Substitute the given values and find the change in volume: $$\mathrm{\Delta }V=1.26L-5.00L=-3.74L$$ The negative sign indicates that the volume decrease occurred.

Convert the pressure from atm to Pa

Given pressure in atm: $P=0.857atm$. We need to convert it to Pascal (Pa), and we can use the following conversion factor: $1atm=101.325kPa=101325Pa$. Now, let's calculate the pressure in Pa: $$P=0.857atm\times 101325Pa/atm=86834.275Pa$$

Calculate the work done

Now we have the pressure in Pa, and we can use the formula for work $W=-P\mathrm{\Delta }V$. Substitute the values (note that $\mathrm{\Delta }V$ should be in m³; 1 L = 0.001 m³). $$W=-86834.275Pa\times (-3.74L)(0.001\frac{m^3}{L})$$

Calculate the final value for work

Perform the calculation to find the work: $$W=86834.275Pa\times 0.00374m^3=324.799J$$ The work involved in the chemical reaction is approximately 324.8 J.

Key concepts

Work-Energy Principle in Chemistry

Understanding the work-energy principle is crucial for studying the behavior of systems in chemistry, especially during chemical reactions or physical processes such as expansion or compression of gases. In simple terms, work is done when a force causes a displacement. In the context of chemistry, we often refer to the work done by or on a system as the energy transferred when a system expands or contracts against an external pressure.

The work-energy principle in chemistry can be translated to the relationship between work and the internal energy of a system. When work is done on a system, such as compressing a gas, energy is transferred to the system, increasing its internal energy. Conversely, when a system does work on its surroundings, for instance by expanding against an external pressure, it loses energy. The formula to calculate this work is given by the equation, $$W=-P\mathrm{\Delta }V$$ where 'W' is the work done, 'P' is the external pressure, and '\Delta V' is the change in volume. The sign convention is such that work is considered positive when done on the system, and negative when the system does work on the surroundings.

Pressure-Volume Work

Pressure-volume work, often abbreviated as P-V work, is one specific type of work encountered in chemical processes and thermodynamics. It occurs when the volume of a system changes in the presence of an external pressure. For gaseous reactions or transitions, this is the most common form of work encountered.

In our exercise, the chemical reaction caused the volume to decrease, implying that the surroundings did work on the system. To calculate this pressure-volume work, you need two critical pieces of information: the external pressure and the change in volume during the process. It's important to note that P-V work is scalar and thus does not depend on the path but only on the initial and final states of the system.

To correctly ascertain the work done, it's essential to align the units of pressure and volume. Ideally, the pressure should be in Pascals (Pa) and the volume in cubic meters (m3). Consistent units ensure an accurate calculation of work in Joules, as was demonstrated in the step-by-step solution of our exercise.

Gas Laws and Work

The behavior of gases during a process that involves a change in volume can be precisely described by the gas laws. These laws, which include Boyle's Law, Charles's Law, and Avogadro's Law, help us predict how gases react to changes in volume, temperature, and pressure. In the realm of work calculation, Boyle's Law is particularly relevant, as it describes the inverse relationship between pressure and volume at a constant temperature for a fixed amount of gas.

Considering Boyle's Law, when a gas is compressed (decreasing volume), its pressure increases if temperature is constant, and similarly, when a gas expands, its pressure decreases. This physical property plays a crucial role in understanding how work is done in terms of the gas laws. If you know the initial and final states of the gas regarding pressure and volume, you can calculate the work performed without knowing the path taken between these states.

In our example, although the gas laws themselves were not explicitly used in the calculation, the underlying concept applies; the volume change dictates the work done, adhering to the principles presented by these fundamental laws.