Question

The volume of \(0.100 \mathrm{~mol}\) of helium gas at \(27^{\circ} \mathrm{C}\) is increased isothermally from \(2.00 \mathrm{~L}\) to \(5.00 \mathrm{~L}\). Assuming the gas to be ideal, calculate the entropy change for the process.

Short answer

The entropy change for the isothermal expansion of the helium gas is approximately \(0.762 \,\mathrm{J\,K^{-1}}\).

Step-by-step solution

List the given information

We have the following information: - \(n = 0.100 \,\mathrm{mol}\) (number of moles of helium gas) - \(T = 27^{\circ} \mathrm{C} = 273.15 + 27 = 300.15 \,\mathrm{K}\) (temperature of the gas) - \(V_i = 2.00 \,\mathrm{L}\) (initial volume of the gas) - \(V_f = 5.00 \,\mathrm{L}\) (final volume of the gas) - Isothermal process (constant temperature) It is also noteworthy that the gas is considered to be ideal. This means the gas obeys the ideal gas law, given by: \[PV = nRT\]

Find the entropy change formula for an isothermal process

For an ideal gas undergoing an isothermal process, the entropy change \(\Delta S\) can be calculated using the following formula: \[\Delta S = nR \ln \frac{V_f}{V_i}\] where - \(n\) is the number of moles of the gas - \(R\) is the gas constant, approximately equal to \(8.314 \,\mathrm{J\,K^{-1}\,mol^{-1}}\) - \(V_i\) and \(V_f\) are the initial and final volumes, respectively

Calculate the entropy change

Now, we can plug in the given values into the formula and calculate the entropy change for the process: \[\Delta S = (0.100 \,\mathrm{mol}) (8.314 \,\mathrm{J\,K^{-1}\,mol^{-1}}) \ln \frac{5.00 \,\mathrm{L}}{2.00 \,\mathrm{L}}\] \[\Delta S = 0.8314 \,\mathrm{J\,K^{-1}} \ln \left(\frac{5}{2}\right)\] \[\Delta S = 0.8314 \,\mathrm{J\,K^{-1}} \times 0.9163\] \[\Delta S \approx 0.762 \,\mathrm{J\,K^{-1}}\] Thus, the entropy change for the isothermal expansion of the helium gas is approximately \(0.762 \,\mathrm{J\,K^{-1}}\).

Key concepts

Ideal Gas Law

The ideal gas law is a fundamental principle used to describe the behavior of gases under various conditions. It is expressed by the equation \(PV = nRT\), where:

• \(P\) is the pressure of the gas.
• \(V\) is the volume of the gas.
• \(n\) is the amount of substance, measured in moles.
• \(R\) is the gas constant, a universal constant that relates energy to temperature and amount of substance.
• \(T\) is the temperature, measured in Kelvin.

The ideal gas law assumes that the gas behaves ideally, meaning that the gas molecules do not interact with each other except through elastic collisions, and occupy negligible volume compared to the volume of the container. This law can be used to relate the thermodynamic variables of a gas and is essential for calculating changes such as those in volume, pressure, or temperature.
The ideal gas law is particularly useful for solving problems involving isothermal, isobaric, isochoric, and adiabatic processes in physics and chemistry.

Isothermal Process

An isothermal process is a type of thermodynamic process in which the temperature remains constant. In such a process, the internal energy of an ideal gas does not change because the temperature is constant. This means that the heat added to the system is completely converted into work done by the gas.
For an isothermal process involving an ideal gas, any changes in volume or pressure occur without a change in temperature, thus the equation \(PV = \,\text{constant}\) holds. This principle is key in understanding how work and heat interchange in isothermal processes.

• During isothermal expansion, as the gas does work on the surroundings by expanding, it absorbs heat to keep the temperature constant.
• During isothermal compression, the gas releases heat to the surroundings since work is done on the gas by compressing it.

In the context of the given problem, this means that when the helium gas expands from 2.00 L to 5.00 L, even though work is being done by the gas, the temperature stays at 300.15 K throughout the entire process.

Helium Gas

Helium is a noble gas, represented by the chemical symbol He, known for its light weight and non-reactive nature. As a monatomic gas, it consists of single atoms and exhibits ideal gas behavior very closely, making it an optimal choice for exercises involving the ideal gas law.

• Helium is colorless, tasteless, and odorless.
• It is the second lightest element and remains in gaseous form at room temperature.
• Because helium atoms do not bond with each other, interactions between particles in a sample of helium gas are minimal. This is why helium behaves nearly ideally, obeying the ideal gas law under a wide range of conditions.

In the context of the problem, helium's ideal behavior allows us to apply theoretical formulas like those based on the ideal gas law, to calculate properties such as entropy change accurately.

Gas Constant

The gas constant \(R\) is an essential part of the ideal gas law, representing the proportionality constant that relates the energy scale to the microscopic scale of molecules. Its value is universally recognized as approximately \(8.314 \, \mathrm{J\,K^{-1}\,mol^{-1}}\). This constant helps in bridging the gap between the macroscopic state-based description of gases and the molecular-level interactions.

• \(R\) is derived from the Boltzmann constant and Avogadro's number, tying together temperature and pressure with the energy impacts on particles of a gas.
• It plays a vital role in thermodynamic equations and helps define relationships in processes involving gases, such as predicting changes in volume or pressure under given temperatures.

In thermodynamics problems, the gas constant efficiently links theoretical models with empirical measurements, facilitating calculations of entropy change and other state functions. For example, in the calculation of the entropy change of helium gas expanding isothermally, \(R\) simplifies to usable values for practical computation.