Question

1.??????(100) Given the van der Waals constants for ethane gas as a = 5.562 L$^2$ bar/mol$^2$, b = 0.06380 L/mol, for 10.0 mol of ethane at 300 K and under 30 bar, where R = 0.08314 L-bar/K-mol (1)find the second virial coefficient B at this temperature. (2)calculate the compressibility factor Z from the first two terms. (3)estimate the approximate molar volume form Z. (4)what is its Boyle temperature $T_B$. (1) $B = b - \frac{a}{RT} = 0.06380 - \frac{5.562}{(0.08314)(300)} = -0.15920$ l/mole $B^{\prime} = \frac{B}{RT} = \frac{0.15920}{0.08314 \times 300} = -6.3827 \times 10^{-3}$ bar$^{-1}$ $(2) Z = 1 + B^{\prime}P = 1 + (-6.3827 \times 10^{-3})(30) = 0.8085$ $\frac{V}{n} = \frac{ZRT}{P} = \frac{(0.8085)(0.08314)(300)}{30} = 0.6722$ l/mole $(4) T_B = \frac{a}{Rb} = \frac{5.562}{(0.08314)(0.06380)} = 1048.6$ K

          1.??????(100)
Given the van der Waals constants for ethane gas as a = 5.562 L$^2$ bar/mol$^2$, b = 0.06380 L/mol,
for 10.0 mol of ethane at 300 K and under 30 bar, where R = 0.08314 L-bar/K-mol
(1)find the second virial coefficient B at this temperature.
(2)calculate the compressibility factor Z from the first two terms.
(3)estimate the approximate molar volume form Z.
(4)what is its Boyle temperature $T_B$.
(1) $B = b - \frac{a}{RT} = 0.06380 - \frac{5.562}{(0.08314)(300)} = -0.15920$ l/mole
$B^{\prime} = \frac{B}{RT} = \frac{0.15920}{0.08314 \times 300} = -6.3827 \times 10^{-3}$ bar$^{-1}$
$(2) Z = 1 + B^{\prime}P = 1 + (-6.3827 \times 10^{-3})(30) = 0.8085$
$\frac{V}{n} = \frac{ZRT}{P} = \frac{(0.8085)(0.08314)(300)}{30} = 0.6722$ l/mole
$(4) T_B = \frac{a}{Rb} = \frac{5.562}{(0.08314)(0.06380)} = 1048.6$ K

1.??????(100)
Given the van der Waals constants for ethane gas as a = 5.562 L^2 bar/mol^2, b = 0.06380 L/mol,
for 10.0 mol of ethane at 300 K and under 30 bar, where R = 0.08314 L-bar/K-mol
(1)find the second virial coefficient B at this temperature.
(2)calculate the compressibility factor Z from the first two terms.
(3)estimate the approximate molar volume form Z.
(4)what is its Boyle temperature TB.
(1) B = b - (a)/(RT) = 0.06380 - (5.562)/((0.08314)(300)) = -0.15920 l/mole
B^' = (B)/(RT) = (0.15920)/(0.08314 × 300) = -6.3827 × 10^-3 bar^-1
(2) Z = 1 + B^'P = 1 + (-6.3827 × 10^-3)(30) = 0.8085
(V)/(n) = (ZRT)/(P) = ((0.8085)(0.08314)(300))/(30) = 0.6722 l/mole
(4) TB = (a)/(Rb) = (5.562)/((0.08314)(0.06380)) = 1048.6 K

Added by Dale S.

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