Aconstant-volume perfect gas thermometer indicates a pressure of 6.69 kpa at the triple point temperature of water (273.16 k). (a) what change of pressure indicates a change of 1.00 k at this temperature? (b) what pressure indicates a temperature of 100.00 °c? (c) what change of pressure indicates a change of 1.00 k at the latter temperature?

(a) ΔP=0.0245 kPa

(b) P=9.14 kPa

(c)ΔP=0.0245 kPa

Explanation:

(a) As it is perfect gas we can use

(P₁V₁)/T₁=(P₂V₂)/T₂

Since this constant volume so

P₁/T₁=P₂/T₂

T₂ is change in temperature

T₂=1.00+273.16

T₂=274.16 K

ΔP=6.71449-6.69

ΔP=0.0245 kPa

(b) As

(c) Same steps as in part (a)

ΔP=9.164-9.14

ΔP=0.0245kPa

4

explanation:

4

double the distance = divide strength by 4