2 moles of a diatomic gas undergoes process : PT₂/V = constant. Then, molar heat capacity of gas d

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11.Thermodynamics

normal

$2$ moles of a diatomic gas undergoes the process : $PT_2/V$ = constant. Then, the molar heat capacity of the gas during the process will be equal to

$5R/2$

$9R/2$

$3R$

$4R$

$: \rightarrow$ Given hrocess is

$\frac{P T^2}{V} =\text { constant }$

$C_v =\frac{5}{2} R \quad \text { For }$

$C_V=\frac{5}{2} R \text { (For diatomic }\text { equation } \text { gas) }$

(We know the gas equation gas)

$P V =n R T$

$\therefore T \propto P V$

$\frac{P P^2 V^2}{V} =\text { constant }$

$P^3 V =k$

$P V^{1 / 3} =\text { constant }-\text { (1) }$

formula for molar heat cahacity

$C =C_v+\frac{R}{1-x}$

$x =1 / 3 \text { from eq (1) }$

$\therefore \quad C =\frac{S}{2} R+\frac{R}{1-\frac{1}{3}}$

$C =4 R$

Standard 11

Physics

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normal

normal

normal

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