The densities of air at 25,^o C and 100,^o C | vedclass

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(B) According to Charles's law,$V \propto T$ at constant pressure. Since density $d = \frac{m}{V}$,we have $d \propto \frac{1}{T}$,which implies $d_1

Vedclass · Accepted Answer

$-200$

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(B) According to Charles's law,$V \propto T$ at constant pressure. Since density $d = \frac{m}{V}$,we have $d \propto \frac{1}{T}$,which implies $d_1 T_1 = d_2 T_2$. Let the absolute zero be $x\,^o C$. Then $T_1 = (25 - x) 	ext{ K}$ and $T_2 = (100 - x) 	ext{ K}$. Given $d_1 = 1.2\ gL^{-1}$ and $d_2 = 0.9\ gL^{-1}$. Substituting the values: $1.2 	imes (25 - x) = 0.9 	imes (100 - x)$. $1.2(25) - 1.2x = 0.9(100) - 0.9x$. $30 - 1.2x = 90 - 0.9x$. $30 - 90 = 1.2x - 0.9x$. $-60 = 0.3x$. $x = \frac{-60}{0.3} = -200\,^o C$.

The densities of air at $25\,^o C$ and $100\,^o C$ are $1.2\ gL^{-1}$ and $0.9\ gL^{-1}$ respectively. Assuming that the air obeys Charles's law,the value of absolute zero of temperature in degree Celsius is .............. $^o C$.

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