Question

The work done during the expansion of a gas from \(4 \mathrm{dm}^{3}\) to \(6 \mathrm{dm}^{3}\) against a constant external pressure of \(3 \mathrm{~atm}\) is \((1 \mathrm{~L} \mathrm{~atm}=101.32 \mathrm{~J})\) (a) \(-6]\) (b) \(-608 J\) (c) \(+304 \mathrm{~J}\) (d) \(-304 \mathrm{~J}\)

Short answer

-608 J

Step-by-step solution

Convert Volume to Liters

First, note that the volume should be in liters for the calculation of work, since the given conversion factor is in liters. As 1 dm^3 is equal to 1 liter, the volume expansion of the gas is from 4 liters to 6 liters.

Calculate Volume Change

Calculate the change in volume which the gas undergoes during expansion. The change in volume \(\Delta V\) is the final volume minus the initial volume. Therefore, \(\Delta V = 6 L - 4 L = 2 L\).

Calculate Work Done

Use the formula for work done by the gas against constant external pressure: \(W = -P_{ext} \Delta V\), where \(P_{ext}\) is the external pressure and \(-\) sign indicates work done by the system on the surroundings. Here, \(P_{ext} = 3 atm\) and \(\Delta V = 2 L\). Convert the work into joules using the given conversion factor (1 L atm = 101.32 J). Calculating gives \(W = -3 atm \times 2 L = -6 L atm = -6 \times 101.32 J = -608.92 J\), which we can round to \(W = -608 J\).

Key concepts

Thermodynamics

Thermodynamics is a branch of physics that deals with heat, work, temperature, and energy transfer. In the context of gas expansion, we are particularly interested in how energy is transferred as work when a gas expands against an external pressure. The laws of thermodynamics govern this process, and understanding them is crucial to calculating the work done.

During an expansion, the internal energy of the gas can do work on its surroundings, often resulting in a volume increase. It's important to remember that in a thermodynamic system, work is defined as the energy transferred by a force acting through a distance. In our case, as the gas expands, it exerts a force on the piston or boundaries containing it, doing work in the process. The direction of work is of significance too: when a system does work on its surroundings, like in our exercise, the work is considered negative, and vice versa.

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in thermodynamics that relates the pressure (P), volume (V), temperature (T), and number of moles (n) of an ideal gas to each other through the equation PV = nRT, where R is the ideal gas constant. While our exercise does not require us to calculate all these variables, it is implicit that the behavior of the gas during its expansion can be described by this law.

This law assumes that the particles of an ideal gas do not interact and take up no space themselves, which simplifies calculations and conceptual understanding, although real gases may exhibit behavior that deviates from this model. The key takeaway from the ideal gas law for solving the exercise is the relationship between pressure and volume, which helps us calculate work during a process where the temperature is implied to be constant (isothermal process).

Pressure-Volume Work

Pressure-volume work, often referred to as P-V work, is the work done by or on a gas as it expands or contracts at a given pressure. The general formula for calculating the work done in an expansion or compression is given by the equation: \( W = -P_{ext} \times \text{Δ}V \), where \( W \) is work, \( P_{ext} \) is the external pressure, and \( \text{Δ}V \) is the change in volume.

This formula tells us that work is directly proportional to both the external pressure and the change in volume. Also, the sign convention is such that work done by the system on the surroundings, as in our exercise, is negative, signifying that energy is leaving the system. To solve the problem from our exercise, the volume change needs to be calculated first, as in Step 2, followed by using the formula to find the work done. Remembering to convert the units properly, as work is traditionally expressed in Joules (J), is fundamental to arriving at the correct answer.