Question

A constant volume perfect gas thermometer indicates a pressure of 6.69 kPaat the triple point of water (273.16 K). (a) What change of pressure indicates a change of 1.0 K at this temperature? (b) What pressure indicates a temperature of 100 ⁰C? (c) What change of pressure indicate a change of 1.0 K at the latter temperature?

Answer 1

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.