The P vs V graph is plotted for 1 mole of a h | vedclass

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(A) For a real gas,the critical temperature is given by $T_c = \frac{8a}{27Rb}$.From the graph,the isotherm at $200 \ K$ shows a horizontal portion,in

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$54 < \frac{a}{b} < 81$

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(A) For a real gas,the critical temperature is given by $T_c = \frac{8a}{27Rb}$.From the graph,the isotherm at $200 \ K$ shows a horizontal portion,indicating that $200 \ K$ is below the critical temperature $(T_c > 200 \ K)$.The isotherm at $300 \ K$ does not show a horizontal portion,indicating that $300 \ K$ is above the critical temperature $(T_c Thus,$200 Substituting $T_c = \frac{8a}{27Rb}$,we get $200 Rearranging for $\frac{a}{b}$,we have $\frac{200 	imes 27 	imes R}{8} Using $R = 0.08$,we get $\frac{200 	imes 27 	imes 0.08}{8} $25 	imes 27 	imes 0.08 $54 < \frac{a}{b} < 81$.

The $P$ vs $V$ graph is plotted for $1$ $mole$ of a hypothetical gas. Based on the given isotherms at $200 \ K$ and $300 \ K$,determine the range of $\frac{a}{b}$ for this gas in $atm-L \ mole^{-1}$. [Given: $R = 0.08 \ atm-L \ mole^{-1} \ K^{-1}$]

Similar Questions

The graph between $P$ and $V$ below the critical temperature is:

Statement $-1$: The compressibility factor at the critical point is the same for different gases following van der Waal's equation.
Statement $-2$: The compressibility factor is independent of pressure at the critical temperature.

At relatively high pressure,the van der Waals equation becomes?

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 $R = 0.083 \ L \ bar \ K^{-1} \ mol^{-1}$)

Under which conditions does a real gas show maximum deviation from ideal gas behavior?

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