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11.Thermodynamics
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સમોષ્મી પ્રક્રિયામાં એક-પરમાણ્વિય વાયુ માટે દબાણ અને તાપમાન માટે $P \propto T^{c}$ છે, તો $c =$.......

A

$0.6$

B

$1.67$

C

$0.4$

D

$2.5$

Solution

સમોષ્મી પ્રક્રિયા માટે, $PV^{\gamma} =$ અચળ

$\therefore \,\,{\text{P}}{\left( {\frac{{\mu {\text{RT}}}}{{\text{P}}}} \right)^\gamma }$ અચળ 

$\therefore \,\,\,{{\text{P}}^{{\text{1 – }}\gamma }}{T^\gamma } = $  અચળ 

$\therefore \,{{\text{P}}^{{\text{1 – }}\gamma }} = \,\,\frac{{અચળ}}{T^\gamma }$

$\,\therefore \,\,\,{{\text{P}}^{{\text{1 – }}\gamma }} = {T^{ – \gamma }}$ (અચળ )

$\therefore \,\,{\text{P}} = $(અચળ )${{\text{T}}^{{\text{ – }}\gamma \left( {\frac{1}{{1 – \gamma }}} \right)}}$

$\therefore \,\,\,P\,\, \propto \,\,{T^{\left( {\frac{\gamma }{{\gamma  – 1}}} \right)}}$ પણ  ${\text{P}}\, \propto \,{T^c}$ આપેલ છે

અચળાંક ${\text{c}} = \frac{\gamma }{{\gamma  – 1}}\,$  એક- પરમાણ્વિય વાયુ માટે,$\gamma  = \frac{5}{3}$

$\therefore \,\,\,c = \frac{{5/3}}{{(5/3) – 1}} = \frac{{5/3}}{{2/3}} = \frac{5}{2}$

 

Standard 11
Physics

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