Question

A gas cylinder is filled with silane (SiH4), used in semiconductor processes. The cylinder’s internal volume is 2.40 L, and it contains 542 g of the compound. Estimate the pressure inside the cylinder at 21oC. The properties of silane: Tc = 269.7 K, Pc = 48.4 bar, and ω = 0.094.

Answer 1

The value is
`P = 7.8 *10^(6) \ Pa`

Step-by-step explanation:

From the question we are told that

The internal volume is
`V_i = 2.40 \ L = 2.40 *10^(-3) \ m^3`

The mass of the compound contained is
`m = 542 \ g`

The temperature is
`T = 21^o C = 21 + 273 = 294 \ K`

The critical temperature of silane is
`T_c = 269.7 \ K`

The critical pressure of silane is
`P_c = 48.4 bar = 48.4 *10^(5) \ Pa`

Generally the number of moles of silane inside the cylinder is mathematically represented as

`n = (m)/(M)`

here M is the molar mass of silane with value
`M = 32 g/mol`

So

`n = (542)/(32)`

=>
`n = 16.4 \ mol`

Generally the molar volume of silane in the cylinder is mathematically represented as

`Vn = (V_i)/(n)`

=>
`Vn = (2.40 *10^(-3))/(16.4)`

=>
`Vn = 1.42*10^(-4) \ m^3 / mol`

Generally from Soave-Redlich-Kwon we have that

`P = (RT)/(V_n - b) - (a)/(V_n (V_n + b))`

Here b is a constant which is mathematically represented as

`b = 0.08664 * (R T_c )/(P_c)`

substituting
`8.314 J/mol\cdot K` for R we have \

`b = 0.08664 * \frac {8.314* 269.7}{48.4*10^(5)}`

`b = 4.0139 *10^(-5) m^3/mol`

a is also a constant which is mathematically represented as

`a = 0.42748 * ((R * T_c)^2)/(P_c) * (1 + m [1-√(T_r) ])^2`

Here
`T_r` is the reduced temperature which is mathematically represented as

`T_r = (T)/(T_c)`

=>
`T_r = 1.09`

m is a constant which is mathematically represented as

`m = 0.480 + 1.574w - 0.176w^2`

=>
`m  =  0.480 + 1.574 (0.094) - 0.176(0.094)^2`

=>
`m = 0.626`

So

`a = 0.42748 * ((8.314 * 269.7)^2)/(48.4*10^(5)) * (1 + 0.626 [1-√(1.09) ])^2`

`a = 0.4198`

From
`P =  (RT)/(V_n - b) - (a)/(V_n (V_n + b))` we have

`P = (8.314 * 294)/(1.42*10^(-4) - 4.0139 *10^(-5) ) - (0.626)/(1.42*10^(-4) (1.42*10^(-4) + 4.0139 *10^(-5) ))`

`P = 7.8 *10^(6) \ Pa`