For a real gas at 25^{circ} C temperature and | vedclass

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(A) For a real gas under high pressure,the compressibility factor $Z$ is given by the equation: $Z = 1 + \frac{Pb}{RT}$.Given $Z = 2$,$P = 99 \ bar$,$

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(A) For a real gas under high pressure,the compressibility factor $Z$ is given by the equation: $Z = 1 + \frac{Pb}{RT}$.Given $Z = 2$,$P = 99 \ bar$,$T = 25 + 273 = 298 \ K$,and $R = 0.083 \ L \ bar \ K^{-1} \ mol^{-1}$.Substituting the values: $2 = 1 + \frac{99 	imes b}{0.083 	imes 298}$.$1 = \frac{99 	imes b}{24.734}$.$b = \frac{24.734}{99} \approx 0.2498 \ L \ mol^{-1}$.$b \approx 0.25 \ L \ mol^{-1} = 25 	imes 10^{-2} \ L \ mol^{-1}$.Thus,the nearest integer value is $25$.

For a real gas at $25^{\circ} C$ temperature and high pressure $(99 \ bar)$,the value of the compressibility factor is $2$. The value of the Van der Waals constant '$b$' is $\times 10^{-2} \ L \ mol^{-1}$. (Nearest integer) (Given $R = 0.083 \ L \ bar \ K^{-1} \ mol^{-1}$)

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