Question

A certain gas has a molecular weight of 30.0, a critical temperature of 310 K, and a critical pressure of 4.5 MPa. Calculate the density in kg / m 3 kg/m3 of this gas at 465 K and 9.0 MPa (a) if the gas is ideal and (b) if the gas obeys the law of corresponding states

Answer 1

Explanation:

(a)  We know that,

           $P (kg/m^{3}) = \frac{m (kg)}{V (m^{3})}$

or,           P = $\frac{(M.W)P}{RT}$

                  = $\frac{30 kg/mol}{465 K} \times \frac{9.0}{0.0826 m^{3}atm/mol K} \times \frac{10 atm}{1.013 kPa}$

                  = 69.8 $kg/m^{3}$  

Hence, if the gas is ideal then its density is 69.8 $kg/m^{3}$.

(b)   It is known that,

           $T_{r} = \frac{T}{T_{c}}$

                       = $\frac{465}{310}$

                       = 1.5

          $P_{r} = \frac{P}{P_{c}}$

                     = $\frac{9}{4.5}$

                     = 2

Expression for the compressibility factor is as follows.

             Z = $\frac{PV}{RT}$

                 = 0.84

  and,       P = $\frac{(M.W)P}{RTZ}$

                    = $\frac{69.8}{0.84}$

                    = 83.095 $kg/m^{3}$

Therefore, if the gas obeys the law of corresponding states then its density is 83.095 $kg/m^{3}$.

Answer 2

The density of the gas at 465 K and 9.0 MPa pressure has been 69.376 kg/$\rm \bold{m^3}$. The density of gas following the law of corresponding states is 82.590 kg/$\rm \bold{m^3}$.

Density can be defined as mass per unit volume. Density can be given by:

Density = $\rm \dfrac{mass}{volume}$

The moles can be calculated by dividing mass by molecular mass.

Moles = $\rm \dfrac{mass}{molecular\;mass}$

Mass = Moles $\times$ Molecular mass

Putting the equation for mass in density:

Density = $\rm \dfrac{Moles\;\times\;Molecular\;mass}{volume}$

$\rm \dfrac{Density}{Molecular\;mass}$ = $\rm \dfrac{Moles}{Volume}$ ......(i)

(a) The gas has been assumed to be an ideal gas. According to the ideal gas equation:

Pressure $\times$ Volume = moles

Pressure = $\rm \dfrac{Moles}{Volume}$ $\times$ Rydberg constant

Putting the values of Moles by Volume from equation (i)

Pressure = $\rm \dfrac{Density}{Molecular\;mass}$ $\times$ Rydberg constant

Density = $\rm \dfrac{Pressure\;\times\;Molecular\;mass}{Rydberg\;Constant\;\times\;Temperature}$

Given:

Pressure = 9.0 MPa = 9.0 $\times$ 9.869 atm = 88.8231 atm

Molecular mass = 30 kg

Rydberg constant = 0.0826 $\rm m^3\;atm/mol\;K$

Temperature = 465 K

Density = $\rm \dfrac{88.8231\;\times\;30}{0.0826\;\times\;465}$

Density = 69.376 kg/$\rm m^3$

The density of the gas at 465 K and 9.0 MPa pressure has been 69.376 kg/$\rm \bold{m^3}$.

(b) The critical temperature of gas = 310 K

The given temperature = 465 K.

The corresponding temperature = $\rm\dfrac{Temperature}{Critical\;temperature}$

The corresponding temperature = $\rm \dfrac{465}{310}$

The corresponding temperature = 1.5 K

The critical pressure = 4.5 MPa

The given pressure = 9.0 MPa

The corresponding pressure = $\rm \dfrac{Pressure}{Critical\;pressure}$

The corresponding pressure = $\rm \dfrac{9.0}{4.5}$

The corresponding pressure = 2 MPa

When the gas follows the corresponding states law:

Z = $\rm \dfrac{Pressure}{Density\;\times\;Rydberg\;constant\;\times\;Temperature}$

The density of the gas following the corresponding states law has been:

Density =  $\rm \dfrac{Density\;of\;Ideal\;gas}{Compresibility\;factor}$

Z = 0.84

Density = 69.376 kg/$\rm \bold{m^3}$.

The density of gas following the law of corresponding states = $\rm \dfrac{69.376}{0.84}$

The density of gas following the law of corresponding states = 82.590 kg/$\rm \bold{m^3}$.

For more information about corresponding state law, refer to the link:

https://brainly.com/question/929044