The pressure of the gas in a constant volume gas thermometer at steam point $(373.15 \mathrm{~K})$ is $1.50 \times 10^4 \mathrm{~Pa}$. What will be the pressure at the triple point of water?

$1.10 \times 10^3 \mathrm{~Pa}$

$1.10 \times 10^2 \mathrm{~Pa}$

$2.10 \times 10^4 \mathrm{~Pa}$

JEE PHYSICS MCQ: THERMAL PROPERTIES OF MATTER

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The pressure of the gas in a constant volume gas thermometer at steam point

(373.15 \mathrm{~K})

1.50 \times 10^4 \mathrm{~Pa}

. What will be the pressure at the triple point of water?

View answer and explanation

Correct option: A

1.10 \times 10^4 \mathrm{~Pa}

Option A (

1.10 \times 10^4 \mathrm{~Pa}

) is correct because it is calculated using the direct proportionality between pressure and temperature in a constant volume gas thermometer.