Could someone please help on just question 2?
1 answer:
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The boiling point of water can be calculated by the equation:
Where:
P = Pressure in mm Hg
Po = Atmospheric pressure in mm Hg
ΔH= heat of vaporization in kJ/mol
R = Ideal Gas Constant (J/mol-K)
To = normal boiling point in Kelvin
T = boiling point of water (K)
Our known values are:
P = 630 mm Hg
Po = 760 mm Hg
ΔH = 40.66 kJ/mol = 40.66×1000 =40660
R = 8.314 J mol⁻¹ K ⁻¹
To = 373 K
Putting these values in the equation,
Solving the equation will give:
T=370K
so, the boiling point of water is 370 K.
Answer:
- 0.07 °C
Explanation:
At constant pressure and number of moles, Using Charle's law
Given ,
V₁ = 439 mL = 0.439 L ( 1 L = 0.001 mL )
V₂ = 0.378 L
T₁ = 317.15 K
T₂ = ?
Using above equation as:
The conversion of T(K) to T( °C) is shown below:
T( °C) = T(K) - 273.15
So, <u>T = 273.08 - 273.15 °C = - 0.07 °C</u>