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: $c_p - c_v = \frac{2.08 	imes 10^7}{4.186 	imes 10^7} \approx 0.5\, cal\, g^{-1} K^{-1}$.

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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