Question

Physics

Topic:-

Thermal expansions


When the bulb of standard gas thermometer is placed in melting ice, the pressure recorded is 83cm of Hg and when immersed in boiling water, it is 113 cm of Hg. When the bulb is placed in another boiling liquid, the pressure recorded is 143 cm of Hg. The boiling point of the liquid is ?

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Answer 1

Answer:

Step-by-step explanation:

Solution :-

Given,

Bulb of standard gas thermometer is placed in melting ice

$\implies t_1=0\degree C$

According to Question :-

$P_1 = 83\:cm\:of\:Hg$

$P_2=143\:cm\:of\:Hg$

Standard gas thermometer volume is constant.

We know that :-

$\beta=\dfrac{1}{273}$

Formula Required :-

$\beta=\dfrac{P_2-P_1}{P_1t_2-P_2t_1}$

$\dfrac{1}{273}=\dfrac{143-83}{83(t_2)-143(0)}$

$\dfrac{1}{273}=\dfrac{60}{83t_2}$

$83t_2=60\times 273$

$83t_2 = 16380$

$t_2=\dfrac{16380}{83}$

$t_2 = 197.34\degree C$