The pressure of the gas in a constant volume gas thermometer is 100 cm of mercury when placed in melting ice at 1 atm. When the bulb is placed in a liquid, the pressure becomes 180 cm of mercury. Temperature of the liquid is :
(Given 0$^\circ$C = 273 K)
Solution
<p>Here volume is constant.</p>
<p>$\therefore$ ${{{P_1}} \over {{T_1}}} = {{{P_2}} \over {{T_2}}}$</p>
<p>$\Rightarrow {{100} \over {273}} = {{180} \over {{T_2}}}$</p>
<p>$\Rightarrow {T_2} = {{180} \over {100}} \times 273$</p>
<p>$= 1.8 \times 273$</p>
<p>$= 491\,K$</p>
About this question
Subject: Physics · Chapter: Thermodynamics · Topic: Zeroth and First Law
This question is part of PrepWiser's free JEE Main question bank. 271 more solved questions on Thermodynamics are available — start with the harder ones if your accuracy is >70%.