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

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,

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

$\displaystyle V = \frac{n\cdot R\cdot T}{P}$.

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:

• $P =\rm 5.00\times 10^{2}\;kPa = 5.00\times 10^{2}\times 10^{3}\; Pa = 5.00\times 10^{5}\; Pa$;
• $V$ 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:

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