If a constant volume gas thermometer records a pressure of 20 kPa at tripple point of water and pressure of 14.3 kPa at the dry ice, then the temperature of dry ice will be

Question

If a constant volume gas thermometer records a pressure of 20 kPa at tripple point of water and pressure of 14.3 kPa at the dry ice, then the temperature of dry ice will be (A)  -77.85° C  (B)  +77.85° C  (C)  -66.05° C  (D) None of these

Tardigrade Admin · Accepted Answer

Given, pT=20.0 kPa and p=14.3 text kPa By the relation, T=(273.16K)( (p/pT) ) =(273.16K)× ( (14.3kPa/20.0kPa) )=195.3K On celsium scale θ (° C)=T(K)-273.15K =195.3-273.15 =-77.85° C

Q. If a constant volume gas thermometer records a pressure of 20 kPa at tripple point of water and pressure of 14.3 kPa at the dry ice, then the temperature of dry ice will be

$- 77.85^{\circ} C$

$+ 77.85^{\circ} C$

$- 66.05^{\circ} C$

None of these

Given, $p_{T} = 20.0 k P a$
and $p = 14.3 k P a$
By the relation, $T = (273.16 K) (\frac{p}{p _{T}})$
$= (273.16 K) \times (\frac{14.3 k P a}{20.0 k P a}) = 195.3 K$
On celsium scale $θ (^{\circ} C) = T (K) - 273.15 K$
$= 195.3 - 273.15$
$= - 77.85^{\circ} C$

Q. If a constant volume gas thermometer records a pressure of 20 kPa at tripple point of water and pressure of 14.3 kPa at the dry ice, then the temperature of dry ice will be

−77.85∘C

+77.85∘C

−66.05∘C

None of these

Given, pT​=20.0kPa and p=14.3kPa By the relation, T=(273.16K)(pT​p​) =(273.16K)×(20.0kPa14.3kPa​)=195.3K On celsium scale θ(∘C)=T(K)−273.15K =195.3−273.15 =−77.85∘C

$- 77.85^{\circ} C$

$+ 77.85^{\circ} C$

$- 66.05^{\circ} C$

Given, $p_{T} = 20.0 k P a$
and $p = 14.3 k P a$
By the relation, $T = (273.16 K) (\frac{p}{p _{T}})$
$= (273.16 K) \times (\frac{14.3 k P a}{20.0 k P a}) = 195.3 K$
On celsium scale $θ (^{\circ} C) = T (K) - 273.15 K$
$= 195.3 - 273.15$
$= - 77.85^{\circ} C$