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.
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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