Answer:
standard entropy of vaporization of ethanol = 142.105 J/K-mol
Explanation:
given data
enthalpy of vaporization of ethanol = 40.5 kJ/mol = 40.5 × 
J/mol
entropy of vaporization of ethanol boiling point = 285 K
to find out
standard entropy of vaporization of ethanol
solution
we get here standard entropy of vaporization of ethanol that is expess as
standard entropy of vaporization of ethanol ΔS = 
.............1
here ΔH is enthalpy of vaporization of ethanol and T is temperature
put value in equation 1
standard entropy of vaporization of ethanol ΔS = 
standard entropy of vaporization of ethanol = 142.105 J/K-mol
Answer:
The new volume of the gas is 276.45 mL.
Explanation:
Charles's law indicates that for a given sum of gas at constant pressure, as the temperature increases, the volume of the gas increases, and as the temperature decreases, the volume of the gas decreases.
Charles's law is a law that mathematically says that when the amount of gas and pressure are kept constant, the quotient that exists between the volume and the temperature will always have the same value:
Analyzing an initial state 1 and a final state 2, it is satisfied:
In this case:
- V1= 250 mL
- T1= 293 K
- V2= ?
- T2= 324 K
Replacing:
Solving:
V2= 276.45 mL
<em><u>The new volume of the gas is 276.45 mL.</u></em>
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