Question

A constant-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?

Answer 1

(a) ΔP=0.0245 kPa

(b) P=9.14 kPa

(c)ΔP=0.0245 kPa

Step-by-step 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

`P_(2)=((6.69)/(273.16) )*274.16\\P_(2)=6.71449 kPa`

ΔP=6.71449-6.69

ΔP=0.0245 kPa

(b) As

`P_(2)=((6.69)/(273.16) )*373.16\\P_(2)=9.14 kPa`

(c) Same steps as in part (a)

`P_(2)=((9.14)/(373.16) )*374.16\\P_(2)=9.164kPa`

ΔP=9.164-9.14

ΔP=0.0245kPa