Question

While resting, the average\({\rm{70 - kg}}\)human male consumes \(14\;{\rm{L}}\) of pure \({{\rm{O}}_2}\) per hour at \({25^\circ }{\rm{C}}\)and \({\rm{100 kPa}}\). How many moles of \({{\rm{O}}_2}\)are consumed by a \({\rm{70\;kg}}\)man while resting for \({\rm{1}}{\rm{.0\;h}}\)?

Short answer

\(0.56\) moles of \({O_2}\) are consumed by a \(70Kg\) man while resting for \(1.0\)h.

Step-by-step solution

Definition of mole

The ideal gas law can be stated as.

\(PV = nRT\)

where,

\(\begin{aligned}{}{\rm{P = pressure of the gas}}{\rm{.}}\\{\rm{V = volume of the gas}}{\rm{. }}\\{\rm{n = number of moles of the gas}}{\rm{. }}\\{\rm{T = temperature of the gas}}{\rm{. }}\end{aligned}\)

and

\(\begin{aligned}{}{\rm{R = Ideal gas constant}}\\{\rm{ = 0}}{\rm{.08206L\;atm mo}}{{\rm{l}}^{{\rm{ - 1}}}}{{\rm{K}}^{{\rm{ - 1}}}}\end{aligned}\)

if the amount of the gas of the value\((n)\)is in moles (mol), the temperature of the value\((T)\)is in kelvins\((K)\) and the pressure \((P)\)is in atmospheres (atm).

Converting pressure in kPa into atm

The given temperatures are in degrees Celsius but not in kelvins (absolute scale).So, we need to convert the temperatures accordingly.

We have

\({0^\circ }C = (0) + 273.15K\)

So,

\(\begin{aligned}{}{25^\circ }C = (25) + 273.15K\\ = 298.15K\end{aligned}\)

Converting pressure in \(kP\)a into atm, we have

\(1{\rm{kPa}} = 0.0098\;{\rm{atm}}\)

So,

\(\begin{aligned}{}100{\rm{kPa}} = (100) \cdot (0.0098){\rm{atm}}.\\ = 0.98\;{\rm{atm}}{\rm{.}}\end{aligned}\)

Therefore, the required pressure is \(0.98{\rm{ }}atm.\)

Calculating number of moles

Consider the converted pressure. We have

\(PV = nRT\)

Number of moles of gas \((n) = \frac{{PV}}{{RT}}\)

\(\begin{aligned}{}n = \frac{{(0.98) \cdot (14)}}{{(0.08206) \cdot (298.15)}}\\n = \frac{{13.75}}{{24.46}}\\n = 0.56{\rm{moles}}\end{aligned}\)

Therefore, \(0.56{\rm{moles}}\)of \({O_2}\) are consumed.