Problem 61 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

Each of the vessels 1 and 2 contain \(1.2\) moles of gaseous helium. The ratio of the vessels volumes is \(V_{2} / V_{1}=2.0\), and the ratio of the absolute temperature of helium in them is \(T_{1} / T_{2}=2.0\). Assuming the gas to be ideal, find the different of gas entropies in these vessels, \(S_{2}-S_{1}\). \((\ln 2=0.7)\) (a) \(0.84 \mathrm{cal} / \mathrm{K}\) (b) \(4.2 \mathrm{cal} / \mathrm{K}\) (c) \(-0.84 \mathrm{cal} / \mathrm{K}\) (d) \(-4.2 \mathrm{cal} / \mathrm{K}\)

Step-by-Step Solution

Answer

-0.84 cal/K

Step 1: State the entropy change formula for an ideal gas

The entropy change for an ideal gas when its volume and temperature change can be given by the formula: \(\Delta S = nR \ln(V_f / V_i) + nC_V \ln(T_f / T_i)\), where \(n\) is the number of moles, \(R\) is the universal gas constant, \(V_i\) and \(V_f\) are the initial and final volumes, \(T_i\) and \(T_f\) are the initial and final temperatures, and \(C_V\) is the molar heat capacity at constant volume.

Step 2: Use the provided temperature and volume ratios

We have been given the volume ratio \(V_{2} / V_{1} = 2.0\) and temperature ratio \(T_{1} / T_{2} = 2.0\). Since we want \(S_{2} - S_{1}\), we can rearrange them as \(V_{f} / V_{i} = V_{2} / V_{1}\) and \(T_{f} / T_{i} = T_{2} / T_{1} = 1/(T_{1} / T_{2})\).

Step 3: Calculate the change in entropy \(S_{2} - S_{1}\)

Plugging the ratios into the entropy change formula, we get \(\Delta S = nR \ln(2) + nC_V \ln(1/2)\). Since \(C_V = (3/2)R\) for a monatomic gas like helium, the equation simplifies to \(\Delta S = S_{2} - S_{1} = 1.2R(\ln(2) - (3/2)\ln(2))\).

Step 4: Substitue the given values and calculate

Using the value \(\ln 2 = 0.7\) and \(R = 1.987 cal/(mol K)\), we can substitute them in the equation: \(\Delta S = 1.2(1.987)(0.7 - (3/2)(0.7))\).

Step 5: Solve for \(S_{2} - S_{1}\)

After performing the calculation, we get \(\Delta S = 1.2(1.987)(-0.35)\), which equals \(-0.84 cal/K\). This corresponds to option (c).

Key Concepts

Entropy Change Formula

When we talk about entropy, we're referring to a measure of the disorder or randomness in a system, and it's a key concept in thermodynamics. The 'entropy change formula' for ideal gases allows us to calculate how much entropy changes when a gas expands, compresses, or changes temperature.

The formula for the change in entropy (\text{\text{\(\text{\)\text{\(\text{\text{\)\text{\text{\(\text{\)\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{/}) for an ideal gas is given by the equation:\text{\(\text{\)\text{\(\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{/})({\text{\text{\)\text{\(\text{\)\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\textν\text\textνννννννννννν\(\text{\)\text{\(\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text{\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\text\textโngโngโngง็งงงงงงงงงงโงโงโงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงโงโงโงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงงง่xΔS = nR \text{ln}(V_f / V_i) + nC_V \text{ln}(T_f / T_i)\)), where:

• \( - \) is the number of moles of the gas,
• \( - \) is the universal gas constant (approximately \( - \) cal/(mol K)),
• \( - \) and \( - \) are the initial and final volumes, respe blockSize=

Molar Heat Capacity

Molar heat capacity, often denoted as \(C_p\) (at constant pressure) or \(C_V\) (at constant volume), is an important property of a substance that indicates how much heat is needed to raise the temperature of a mole of the substance by one degree Celsius (or Kelvin as temperature changes are the same in both scales).

For ideal gases, molar heat capacities are particularly simple to calculate given that they depend on the gas's atomicity: monatomic, diatomic, or polyatomic. Helium is a monatomic gas, which means it consists of single atoms. A unique aspect of monatomic gases is their molar heat capacity at constant volume (\(C_V\)), which is \(\text{\)3/2$$}\( times the universal gas constant (\)R\(), or \)\text{\((3/2)R$$}\). Knowing \(C_V\) allows us to understand how the entropy of an ideal gas changes with temperature.

When we include molar heat capacity in the entropy change formula, as seen in the step-by-step solution provided for the exercise, it enables us to account for the heat transfer related to a substance's change in temperature, which contributes to the overall entropy change.

Volume and Temperature Ratios

The ratios of volume and temperature play a crucial role when we're dealing with gases and their behavior under different conditions. In the context of the given problem, understanding these ratios enables us to apply the entropy change formula effectively.

For ideal gases, Boyle's Law and Charles's Law explain how volume and temperature are related to pressure, provided that the amount of gas and pressure are held constant. In our problem, the ratio of the volumes is \(V_{2} / V_{1} = 2.0\), indicating that the second vessel's volume is twice that of the first one. Similarly, the ratio of the temperatures is \(T_{1} / T_{2} = 2.0\), meaning that the temperature of the gas in the first vessel is twice that of the second vessel.

The inverse relationship between the volume and temperature ratios is key to solving entropy-based problems as it helps us find how an entropy change is affected when the system moves from one state to another. The larger the difference in ratios, the greater the potential change in entropy. By substituting these given ratios into the entropy change formula, we can then solve for the change in entropy between two states of a system, as was demonstrated in the exercise.