Problem 60 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

The entropy change of \(2.0\) moles of an ideal gas whose adiabatic exponent \(\gamma=1.50\), if as a result of a certain process, the gas volume increased \(2.0\) times while the pressure dropped \(4.0\) times, is \((\ln 2=0.7)\) (a) \(-11.64 \mathrm{~J} / \mathrm{K}\) (b) \(+11.64 \mathrm{~J} / \mathrm{K}\) (c) \(-34.92 \mathrm{~J} / \mathrm{K}\) (d) \(+34.92 \mathrm{~J} / \mathrm{K}\)

Step-by-Step Solution

Answer

+11.64 J/K

Step 1: Understand the entropy change for an ideal gas

The entropy change (ΔS) for an ideal gas can be calculated using the equation ΔS=nC_vln(T_2/T_1)+nRln(V_2/V_1), where C_v is the molar heat capacity at constant volume, R is the universal gas constant (8.314 J/(mol K)), T_1 and T_2 are the initial and final temperatures, respectively, and V_1 and V_2 are the initial and final volumes, respectively.

Step 2: Determine the molar heat capacity at constant volume (C_v)

To find C_v, use the relation C_v=R/(γ−1). Plugging in the values, we get C_v=8.314 J/(mol K)/(1.5−1)= 8.314 J/(mol K)/0.5 = 16.628 J/(mol K).

Step 3: Express the temperature in terms of pressure and volume

For an ideal gas, knowing that P_1V_1^{γ}=P_2V_2^{γ} and substituting the given ratios (V_2=2V_1 and P_1=4P_2), we find that T_2/T_1=(V_2/V_1)^{γ−1}=(2)^{1.5−1}=2^{0.5}=√2.

Step 4: Calculate the entropy change (ΔS)

Now we can find ΔS using the formula, remembering to substitute √2 for the temperature ratio and using ln(√2)=ln(2)/2=0.7/2=0.35. ΔS=2.0 mol × 16.628 J/(mol·K) × ln(√2) + 2.0 mol × 8.314 J/(mol·K) × ln(2). After calculating, ΔS=2.0 × 16.628 × 0.35 + 2.0 × 8.314 × 0.7=11.64 J/K.

Key Concepts

Adiabatic Process

An adiabatic process is a thermodynamic process in which there is no heat transfer into or out of the system. The system can do work on or by the environment, but the exchange of heat is strictly prohibited during the transition. In other words, during an adiabatic process, the system's heat, symbolized by Q, remains constant, often expressed as: \( Q = 0 \).To understand this concept better, imagine a piston filled with an ideal gas. If the piston is compressed or expanded very quickly, there might not be enough time for heat to enter or leave the gas, creating an adiabatic scenario. Mathematically, the adiabatic process is also characterized by the adiabatic exponent, denoted as \(\gamma\), which is the ratio of the molar heat capacity at constant pressure \(C_p\) to the molar heat capacity at constant volume \(C_v\), given by \(\gamma = \frac{C_p}{C_v}\). For an ideal gas undergoing an adiabatic process, pressure and volume are related by the equation \(P_1V_1^{\gamma} = P_2V_2^{\gamma}\). This relationship is crucial in understanding how temperature, pressure, and volume change with respect to each other without heat exchange.

Molar Heat Capacity

Molar heat capacity is a physical property that indicates the amount of heat needed to raise the temperature of one mole of a substance by one degree Celsius at constant volume or pressure. It's an intrinsic characteristic of materials, and it's represented by two specific values depending on the conditions: \(C_v\) for constant volume and \(C_p\) for constant pressure.

Relation to the Ideal Gas Law

For an ideal gas, these capacities are related through the equation \(C_p = C_v + R\), where R is the ideal gas constant. In the exercise, we look at the molar heat capacity at constant volume, \(C_v\), which is used when a gas is kept at the same volume but allowed to change temperature. This value is crucial for determining entropy change, as entropy is a measure of the disorder or randomness of the particles in a system, and heat capacity directly impacts the change in temperature and thus the change in entropy.The relationship between molar heat capacity and the adiabatic exponent \(\gamma\) is given by \(C_v = \frac{R}{\gamma - 1}\), which allows one to calculate \(C_v\) when \(\gamma\) is known, as in the provided exercise.

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry and physics that describes the behavior of an ideal gas, and is shown as \(PV = nRT\). In this relationship:

• \(P\) is the pressure of the gas,
• \(V\) is the volume of the gas,
• \(n\) is the amount of substance in moles,
• \(R\) is the ideal gas constant,
• \(T\) is the temperature of the gas in Kelvin.

This equation shows that under certain conditions of temperature and pressure, the volume of a gas is directly proportional to the number of moles and temperature, and inversely proportional to the pressure. The Ideal Gas Law is especially helpful because it allows us to relate all these different physical quantities together. When any three of the four variables (P, V, n, T) are known, the fourth can be calculated.

Entropy and the Ideal Gas Law

In the context of entropy, the Ideal Gas Law helps in understanding how gas molecules spread out or become confined and how these volume changes impact the randomness or disorder within a gas sample. The equation mentioned in the exercise to calculate entropy change incorporates both the volume and temperature changes, therefore connecting the Ideal Gas Law directly to the concept of entropy.