Question

The pressure of the gas in a constant volume gas thermometer at steam point (373.15K) is $1.5\times {10}^{4}Pa$. What will be the pressure at the triple point of water ?

• A. $1.10\times {10}^{4}Pa$
• B. $1.50\times {10}^{4}Pa$
• C. $2.10\times {10}^{4}Pa$
• D. $2.50\times {10}^{4}Pa$

Answer 1

The correct option is A $1.10\times {10}^{4}Pa$

$\frac{P}{T}=\frac{P_{tm}}{T_{tr}};$ we know $T_{tr}=273.16K$
Here given $P=1.5\times {10}^{4}Pa;T=373.15K$
$P_{tr}=\frac{1.5\times {10}^4\times 273.16}{373.15}$
$P_{tr}=1.10\times {10}^{4}Pa$