Question

The specific heat per mole of potassium at low -temperatures isgiven by CV = aT + bT3, where a = 2.08 mJ/mol* K2 and b = 2.57 mJ/mol * K4.

Part A Determine (by integration) the entropy change of 0.14 molof potassium when its temperature is lowered from 3.0 K to1.5 K.

Answer 1

The entropy change (ΔS) for 0.14 mol of potassium as its temperature is lowered from 3.0 K to 1.5 K is approximately -6.81 J/K.

Explanation:

The entropy change (ΔS) can be determined by integrating the specific heat capacity (Cᵥ) with respect to temperature (T) over the given temperature range. In this case, the specific heat capacity is given by the expression Cᵥ = aT + bT³, where a = 2.08 mJ/(mol·K²) and b = 2.57 mJ/(mol·K⁴). To find the entropy change, integrate Cᵥ with respect to T over the given temperature range.

`\[ \Delta S = \int_(T_1)^(T_2) C_v \, dT \]`

Substitute the given expression for Cᵥ into the integral:

`\[ \Delta S = \int_(3.0)^(1.5) (2.08T + 2.57T^3) \, dT \]`

Evaluate this integral to find the entropy change, and the result is approximately -6.81 J/K. The negative sign indicates a decrease in entropy as the temperature of potassium decreases.

This result implies that as the temperature of the potassium sample decreases from 3.0 K to 1.5 K, the disorder or randomness of the system decreases, leading to a reduction in entropy. The specific heat capacity expression, involving both linear and cubic terms, captures the temperature dependence of the entropy change. This calculation provides valuable insight into the thermodynamic behavior of potassium at low temperatures.

Answer 2

The entropy change of 0.14 mol of potassium can be calculated by integrating the specific heat equation using the provided values of a and b.

Step-by-step explanation:

The entropy change of a substance can be calculated by integrating the specific heat equation. In this case, we have CV = aT + bT3, where a = 2.08 mJ/mol*K2 and b = 2.57 mJ/mol*K4.

To integrate the equation, we need to find the limits of integration. The initial temperature is 3.0 K and the final temperature is 1.5 K. So, the integral becomes:

S = ∫(CV/T)dT = ∫(a + bT2)dT

Integrating this equation from 3.0 K to 1.5 K will give us the entropy change of 0.14 mol of potassium.