Question

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?

Answer 1

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

Step-by-step explanation:

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

By the ideal gas law,

`P\cdot V = n\cdot R\cdot T`,

`\displaystyle V = (n\cdot R\cdot T)/(P)`.

where

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

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

• 
  `P =\rm 5.00* 10^(2)\;kPa = 5.00* 10^(2)* 10^(3)\; Pa = 5.00* 10^(5)\; Pa`;
• 
  `V` shall be in cubic meters,
  `\rm m^(3)`;
• 
  `T = \rm 52.0 \textdegree C = (52.0 + 273.15)\; K = 325.15\; K`.

Apply the ideal gas law:

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