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11.Thermodynamics
normal

$\beta = - (dV/dP)/V$  સમીકરણ માટે અચળ તાપમાન માટે વિરુધ્ધ $P$ નો ગ્રાફ

A
B
C
D

Solution

$PV = constant $==> $PdV + VdP = 0$ ==> $ – \frac{1}{V}\left( {\frac{{dV}}{{dP}}} \right) = \frac{1}{P}$

$\beta = \frac{1}{P}$   graph will be rectangular hyperbola.

Standard 11
Physics

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