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an engineer wishes to design a container that will hold 12.0 mol of ethane at a pressure no greater than 5.00x10*2 kPa and a temperature of 52.0 degrees celsius. what is the minimum volume the container can have?

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jacob193

Answer:

The minimum volume of the container is 0.0649 cubic meters, which is the same as 64.9 liters.

Explanation:

Assume that ethane behaves as an ideal gas under these conditions.

By the ideal gas law,

[tex]P\cdot V = n\cdot R\cdot T[/tex],

[tex]\displaystyle V = \frac{n\cdot R\cdot T}{P}[/tex].

where

  • [tex]P[/tex] is the pressure of the gas,
  • [tex]V[/tex] is the volume of the gas,
  • [tex]n[/tex] is the number of moles of particles in this gas,
  • [tex]R[/tex] is the ideal gas constant, and
  • [tex]T[/tex] is the absolute temperature of the gas (in degrees Kelvins.)

The numerical value of [tex]R[/tex] will be [tex]8.314[/tex] if [tex]P[/tex], [tex]V[/tex], and [tex]T[/tex] are in SI units. Convert these values to SI units:

  • [tex]P =\rm 5.00\times 10^{2}\;kPa = 5.00\times 10^{2}\times 10^{3}\; Pa = 5.00\times 10^{5}\; Pa[/tex];
  • [tex]V[/tex] shall be in cubic meters, [tex]\rm m^{3}[/tex];
  • [tex]T = \rm 52.0 \textdegree C = (52.0 + 273.15)\; K = 325.15\; K[/tex].

Apply the ideal gas law:

[tex]\displaystyle \begin{aligned}V &= \frac{n\cdot R\cdot T}{P}\\ &= \frac{12.0\times 8.314\times 325.15}{5.00\times 10^{5}}\\ &= \rm 0.0649\; m^{3} \\ &= \rm (0.0649\times 10^{3})\; L \\ &=\rm 64.9\; L\end{aligned}[/tex].

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