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11.Thermodynamics
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$\beta = - (dV/dP)/V$ 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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