This problem is to use the Claussius-Clapeyron Equation, which is:
ln [p2 / p1] = ΔH/R [1/T2 - 1/T1]
Where p2 and p1 and vapor pressure at estates 2 and 1
ΔH is the enthalpy of vaporization
R is the universal constant of gases = 8.314 J / mol*K
T2 and T1 are the temperatures at the estates 2 and 1.
The normal boiling point => 1 atm (the pressure of the atmosphere at sea level) = 101,325 kPa
Then p2 = 101.325 kPa
T2 = ?
p1 = 54.0 kPa
T1 = 57.8 °C + 273.15K = 330.95 K
ΔH = 33.05 kJ/mol = 33,050 J/mol
=> ln [101.325/54.0] = [ (33,050 J/mol) / (8.314 J/mol*K) ] * [1/x - 1/330.95]
=> 0.629349 = 3975.22 [1/x - 1/330.95] = > 1/x = 0.000157 + 1/330.95 = 0.003179
=> x = 314.6 K => 314.6 - 273.15 = 41.5°C
Answer: 41.5 °C
The cells get into your body by the lungs this answer is correct cause i had the same question and my teacher mark it correct
Answer:
The correct answer is B. It is spontaneous only at low temperatures.
Explanation:
In thermodynamics, the Gibbs free energy is a thermodynamic potential that can be used to calculate the maximum of reversible work that may be performed by a thermodynamic system at a constant temperature and pressure.
The spontaneity of a reaction is given by the equation:
ΔG = ΔH - TΔS
where:
ΔH: enthalpy variation
T: absolute temperature
ΔS: entropy variation
As the reaction is exothermic, ΔH<0
As the reaction order increases (the reagents are solid and gas and their product is solid), ΔS<0
Therefore, the reaction will be spontaneous when ΔG is negative.
ΔG = ΔH - TΔS
That is, the entropy term must be smaller than the enthalpy term.
Hence, the reaction will be spontaneous only at low temperatures.
Its d its the only one that makes sense
