Answer:
The value is [tex]P = 7.8 *10^{6} \ Pa [/tex]
Explanation:
From the question we are told that
The internal volume is [tex]V_i = 2.40 \ L = 2.40 *10^{-3} \ m^3[/tex]
The mass of the compound contained is [tex]m = 542 \ g[/tex]
The temperature is [tex]T = 21^o C = 21 + 273 = 294 \ K[/tex]
The critical temperature of silane is [tex]T_c = 269.7 \ K[/tex]
The critical pressure of silane is [tex]P_c = 48.4 bar = 48.4 *10^{5} \ Pa[/tex]
Generally the number of moles of silane inside the cylinder is mathematically represented as
[tex]n = \frac{m}{M}[/tex]
here M is the molar mass of silane with value [tex]M = 32 g/mol[/tex]
So
[tex]n = \frac{542}{32}[/tex]
=> [tex]n = 16.4 \ mol[/tex]
Generally the molar volume of silane in the cylinder is mathematically represented as
[tex]Vn = \frac{V_i}{n}[/tex]
=> [tex]Vn = \frac{2.40 *10^{-3}}{16.4}[/tex]
=> [tex]Vn = 1.42*10^{-4} \ m^3 / mol[/tex]
Generally from Soave-Redlich-Kwon we have that
[tex]P = \frac{RT}{V_n - b} - \frac{a}{V_n (V_n + b)}[/tex]
Here b is a constant which is mathematically represented as
[tex]b = 0.08664 * \frac{R T_c }{P_c}[/tex]
substituting [tex]8.314 J/mol\cdot K[/tex] for R we have \
[tex]b = 0.08664 * \frac {8.314* 269.7}{48.4*10^{5}}[/tex]
[tex]b = 4.0139 *10^{-5} m^3/mol[/tex]
a is also a constant which is mathematically represented as
[tex]a = 0.42748 * \frac{(R * T_c)^2}{P_c} * (1 + m [1-\sqrt{T_r} ])^2[/tex]
Here [tex]T_r[/tex] is the reduced temperature which is mathematically represented as
[tex]T_r = \frac{T}{T_c}[/tex]
=> [tex]T_r = 1.09[/tex]
m is a constant which is mathematically represented as
[tex]m = 0.480 + 1.574w - 0.176w^2[/tex]
=> [tex]m = 0.480 + 1.574 (0.094) - 0.176(0.094)^2[/tex]
=> [tex]m = 0.626[/tex]
So
[tex]a = 0.42748 * \frac{(8.314 * 269.7)^2}{48.4*10^{5}} * (1 + 0.626 [1-\sqrt{1.09} ])^2[/tex]
[tex]a = 0.4198[/tex]
From [tex]P = \frac{RT}{V_n - b} - \frac{a}{V_n (V_n + b)}[/tex] we have
[tex]P = \frac{8.314 * 294}{1.42*10^{-4} - 4.0139 *10^{-5} } - \frac{0.626}{1.42*10^{-4} (1.42*10^{-4} + 4.0139 *10^{-5} )}[/tex]
[tex]P = 7.8 *10^{6} \ Pa [/tex]
