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Crazy boy [7]
3 years ago
8

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?
Physics
1 answer:
mina [271]3 years ago
8 0

Answer:

(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

P_{2}=(\frac{6.69}{273.16} )*274.16\\P_{2}=6.71449 kPa

ΔP=6.71449-6.69

ΔP=0.0245 kPa

(b) As

P_{2}=(\frac{6.69}{273.16} )*373.16\\P_{2}=9.14 kPa

(c) Same steps as in part (a)

P_{2}=(\frac{9.14}{373.16} )*374.16\\P_{2}=9.164kPa

ΔP=9.164-9.14

ΔP=0.0245kPa

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