Respuesta :

Answer:

At 31.88 atm pressure of ethane will have density of 37.2 g/L at 40.0 °C.

Explanation:

To calculate the pressure of gas, we use the equation given by ideal gas equation:

[tex]PV=nRT[/tex]

Number of moles (n)

can be written as: [tex]n=\frac{m}{M}[/tex]

where, m = given mass

M = molar mass

[tex]PV=\frac{m}{M}RT\\\\PM=\frac{m}{V}RT[/tex]

where,

[tex]\frac{m}{V}=d[/tex] which is known as density of the gas

The relation becomes:

[tex]PM=dRT[/tex]

Where :

P = pressure of the gas

R = universal gas constant

T = temperature of the gas

We are given with:

Density of the gas = d = 37.2 g/L

Molar mass of the ethane gas = M = 30 g/mol

Temperature of the ethane gas = T = 40.0°C= 313.15 K

Pressure of the ethane gas = P

[tex]P=\frac{dRT}{M}=\frac{37.2 g/L\times 0.0821 atm L/mol K\times 313.15 K}{30 g/mol}[/tex]

P = 31.88 atm

At 31.88 atm pressure of ethane will have density of 37.2 g/L at 40.0 °C.

Lanuel

The pressure of an ethane gas with a density of 37.2 g/L at 40.0 °C is equal to 31.87 atm.

Given the following data:

  • Density = 37.2 g/L
  • Temperature = 40.0 °C

Molar mass ([tex]C_2H_6[/tex]) = [tex](12 \times 2 + 1 \times 6) = (24+6) =30 \;g/mol[/tex]

Ideal gas constant, R = 0.0821L⋅atm/mol⋅K

Conversion:

Temperature = 40.0 °C to K = [tex]273 +40=[/tex] 313K

To find the pressure of ethane gas, we would use the ideal gas law equation;

[tex]PV = nRT[/tex]

Where;

  • P is the pressure.
  • V is the volume.
  • n is the number of moles of substance.
  • R is the ideal gas constant.
  • T is the temperature.

[tex]Density = \frac{Mass}{Volume}[/tex]

In terms of density, the ideal gas law equation becomes:

[tex]Density = \frac{M_MP}{RT}[/tex]

Where;

  • [tex]M_M[/tex] is the molar mass.

Making P the subject of formula, we have;

[tex]P = \frac{DRT}{M_M}[/tex]

Substituting the given parameters into the formula, we have;

[tex]P = \frac{37.2 \times 0.0821 \times 313}{30}\\\\P = \frac{955.94}{30}[/tex]

Pressure, P = 31.87 atm.

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