(a) Show that the work done by an ideal gas during the quasi-static, isothermal expansion from an initial pressure \(P_{i}\) to a final pressure \(P_{f}\) is given by $$ W=n R T \ln \frac{P_{f}}{P_{i}} $$ (b) Calculate the work done when the pressure of 1 mol of an ideal gas is decreased quasi-statically from 20 to \(1 \mathrm{~atm}\), the temperature remaining constant at \(20^{\circ} \mathrm{C}\) \((R=8.31 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{deg})\)

To understand the work done by an ideal gas during expansion, we start with the basic formula for work in thermodynamics. Work (W) done by a gas during an expansion process is given by the integral of pressure (P) with respect to volume (V):
\[ W = \int_{V_i}^{V_f} P \, dV \]
During an isothermal process (constant temperature), the work done by an ideal gas can be specifically figured out using the ideal gas law (PV=nRT) because the pressure is a function of volume. By integrating this equation, we derive the formula for work done by an ideal gas in an isothermal process:
\[ W = nRT \ln\frac{V_f}{V_i} \]

In an isothermal process, the temperature of the gas remains constant throughout. This is an important condition because it directly impacts how we calculate work.

For an ideal gas, the relationship between pressure, volume, and temperature is governed by the ideal gas law (PV=nRT). Because temperature (T) stays constant in an isothermal process, the product of pressure and volume must also remain constant. This leads to the equation:
\[ P_i V_i = P_f V_f \]
Where P_i and V_i are the initial pressure and volume and P_f and V_f are the final pressure and volume, respectively.

Using this relationship, we can substitute the volume ratio (V_f / V_i) with the inverse pressure ratio (P_i / P_f) in the work done formula.

The ideal gas law is fundamental to understanding how gases behave. Simply stated, it relates the pressure (P), volume (V), temperature (T), and the number of moles (n) of a gas through the equation:
\[ PV = nRT \]
Here, R is the universal gas constant, which has a value of 8.31 J/mol·K.

This law is crucial for deriving other important equations, like the formula for work done in isothermal processes. Because it assumes the behavior of the gas is ideal (perfectly elastic collisions, no intermolecular forces, etc.), we can simplify many complex real-world gas behaviors into more manageable mathematical problems.

A quasi-static process is one which occurs infinitely slowly, ensuring the system is in a state of equilibrium at every stage. This means the process is reversible, and there is no sudden change in pressure or volume.

For an isothermal expansion, a quasi-static process allows us to accurately calculate the work done by the gas since we can assume the gas remains in thermodynamic equilibrium throughout the process. This assumption makes our derivation feasible and ensures we can use the integration method with reliable results.
By understanding that a quasi-static process ensures every intermediate step is balanced, we can apply basic thermodynamic laws confidently to calculate various properties like work.