The difference between the two specific heats | vedclass

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(D) For $1\, g$ of helium gas,the specific gas constant $r$ is given by $r = \frac{R}{M_w}$.Using the ideal gas equation $PV = nRT$,where $n = \frac{m

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$0.5$

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(D) For $1\, g$ of helium gas,the specific gas constant $r$ is given by $r = \frac{R}{M_w}$.Using the ideal gas equation $PV = nRT$,where $n = \frac{m}{M_w}$,we have $r = \frac{PV}{mT}$.At $NTP$,$P = 76\, cm$ of $Hg = 76 	imes 13.6 	imes 981\, dyne/cm^2$,$V = 22400\, cm^3$ (volume of $4\, g$ of He),$m = 4\, g$,and $T = 273\, K$.Thus,$r = \frac{76 	imes 13.6 	imes 981 	imes 22400}{4 	imes 273} \approx 2.08 	imes 10^7\, erg\, g^{-1} K^{-1}$.According to Mayer's relation,$c_p - c_v = \frac{r}{J}$.Substituting the values: $c_p - c_v = \frac{2.08 	imes 10^7}{4.186 	imes 10^7} \approx 0.5\, cal\, g^{-1} K^{-1}$.

Molecule type	$\frac{C_{P}}{C_{v}}$
$A$. Monoatomic	$I$. $\frac{7}{5}$
$B$. Diatomic rigid molecules	$II$. $\frac{9}{7}$
$C$. Diatomic non-rigid molecules	$III$. $\frac{4}{3}$
$D$. Triatomic rigid molecules	$IV$. $\frac{5}{3}$

	Gases		$C_P/C_V$ value
$A$	Monoatomic	$I$	$(4+f)/(3+f)$
$B$	Diatomic (rigid)	$II$	$5/3$
$C$	Diatomic (non-rigid)	$III$	$7/5$
$D$	Polyatomic	$IV$	$9/7$

The difference between the two specific heats of $1\, g$ of helium gas at $NTP$ is .... $cal\, g^{-1} K^{-1}$. (Atomic weight of helium $= 4$ and $J = 4.186 \times 10^7\, erg\, cal^{-1}$)

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Gases $C_P/C_V$ value

$A$ Monoatomic $I$ $(4+f)/(3+f)$

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$C$ Diatomic (non-rigid) $III$ $7/5$

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