1. The Hvap of a certain compound is 39.46 kJ·mol–1 and its Svap is 58.87 J·mol–

Question

1. The Hvap of a certain compound is 39.46 kJ·mol–1 and its Svap is 58.87 J·mol–1·K–1. What is the boiling point of this compound?

2. The molar heat capacity for carbon monoxide at constant volume is CV,m = 20.17 J/(K·mol). A 17.00-L fixed-volume flask contains CO(g) at a pressure of 5.00 kPa and a temperature of 25.0 °C. Assuming that carbon monoxide acts as an ideal gas and that its heat capacity is constant over the given temperature range, calculate the change in entropy for the gas when it is heated to 800.0 °C

3.At very low temperatures, heat capacity, C, is directly proportional to T3 for most substances. Find an expression for the absolute molar entropy, S, at a low temperature T, in terms of C.

4. Consider a process in which an ideal gas is compressed to one-third of its original volume at constant temperature. Calculate the entropy change per mole of gas.

Explanation / Answer

Answer-1

Enthalpy of vaporization (Hvap), ENtropy of vaporization (Svap), Boilinig point (Tb) is related through equation,

Hvap = Tb x Svap

Given data:

Hvap = 39.46 kJ·mol–1 = 39460 J·mol–1

Svap = 58.87 J·mol–1·K–1.

Using above equation,

Tb = 39460 / 58.87

Tb = 670.3 K

Boiling point of this compound is 670.3 K

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