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# Volume coefficient according to charle's law is

$\begin {array} {1 1}(a)\;Change\; in \;Volume\; of\; gas\; for\; each\; 1^{\large\circ} \;rise\; in\; temperature\\(b)\;3.661\times10^{-3}C^{-1}\\(c)\;\large\frac{V-V_{\large\circ}}{tV_{\large\circ}}\\(d)\;Either \;of\; these \end {array}$

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A)
Volume coefficient $\propto$ is given as
$V_{\large\circ} = K\times273.15$
$V_t = K\times(t+273.15)$
$\therefore \large\frac{V_t}{V_{\large\circ}} = \large\frac{t+273.15}{273.15}$
$=[1+\large\frac{t}{273.15}]$
$Or\; V_t = V_{\large\circ} [1+\alpha t]$
Where $\alpha = \large\frac{1}{273} = 3.66\times10^{-3}$
Hence answer is (b)