Answer:
(a) = 6.714 kPa
(b) = 9.14 kPa
(c) = 2.47 kPa
Explanation:
Pressure Law: Pressure law states that the pressure of a fixed mass of gas is directly proportional to its absolute temperature provided the volume is kept constant. It is expressed mathematically as
P₁/T₁ = P₂/T₂
Triple point of water: The triple point of water is the temperature at which water is in equilibrium with the three phases ( solid, liquid and gaseous state.)
Applying Pressure law,
P₁/T₁ = P₂/T₂ .......................... Equation 2
Where P₁ = Initial pressure, T₁ = initial Temperature, P₂ = final Temperature, T₂ = Final Temperature.
(a) At 1.0 K change in temperatures
Making P₂ The subject of the equation in Equation 2,
P₁ = 6.69 kPa = 6.69 × 10³ Pa, T₁ = 273.16 K, T₂ = 273.16 + 1 (change in temperature of 1 .0 K) = 274.16 K.
Substituting these values into equation 2
P₂ =( P₁ × T₂)/T₁
P₂ = (6.69 × 274.16)/273.16
P₂ = 1834.13/273.16
P₂ =6.714 kPa
(b) At a temperature of 100 ⁰C.
P₂ = ( P₁ × T₂)/T₁
Where P₁ = 6.69 Kpa, T₁ = 273.16 K, T₂ = 100 + 273 = 373 K
Substituting these values into the equation above,
∴ P₂ = (6.69×373)/273.16
P₂ = 2495.37/273.16
P₂ = 9.14 kPa
(c)P₂ = ( P₁ × T₂)/T₁
Where, P₁ = 6.69 Kpa, T₁ = 273.16K, T₂ = 373 + 1 = 374 K
∴ P₂ = (6.69 × 374)/273.16
P₂ = 2502.06/273.16
P₂ = 9.16 kPa.
∴change in pressure (ΔP) = P₂ - P₁
(ΔP) = 6.69 - 9.16 = 2.47 kPa.
