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# The critical constant for water are $374^{\large\circ}C$ 218 atm and 0.0566 litre $mol^{-1}$. Calculate a , b and R.

$\begin {array} {1 1}(a)\;0.0189\;litre\;mol^{-1},2.095\;litre^2atm,0.05086\;litre\;atm\;K^{-1}\;mol^{-1}\\(b)189\;litre\;mol^{-1},0.95\;litre^2atm,0.05086\;litre\;atm\;K^{-1}\;mol^{-1}\\(C)\;0.0189\;litre\;mol^{-1},20.95\;litre^2atm,5.086\;litre\;atm\;K^{-1}\;mol^{-1}\\(d)none\;of\;these \end {array}$

Can you answer this question?

Given
$T_C = 374^{\large\circ}C$
= 374 + 273 = 647 K
$P_C = 218 \;atm$
$V_C = 0.0566\;litre\;mol^{-1}$
$b = \large\frac{V_C}{3} = \large\frac{0.0566}{3}$
$=0.0189 \;litre\;mol^{-1}$
$a = 3P_CV^2_C = 3\times218\times(0.0566)^2$
$= 2.095\;litre^2\;atm$
$R = \large\frac{8}{3} \large\frac{P_CV_C}{T_C} = \large\frac{8\times218\times0.0566}{3\times647}$
$= 0.05086 \;litre\;atm\;K^{-1}\;mol^{-1}$
Hence answer is (a)
answered Apr 2, 2014