91影视

A substance's volume is the amount of space it occupies.

Let us solve the given problem.

We have,

\(PV = nRT\)

where,

\(P = \)pressure of the gas.

\(V = \)volume of the gas.

\(n = \)number of moles of the gas.

\(T = \)temperature of the gas.

\(R = \)Ideal gas constant

\( = 0.08206L{\rm{ atm mo}}{{\rm{l}}^{ - 1}}{K^{ - 1}}\)

If, amount of the gas \((n)\)is in moles, temperature \((T)\) is in Kelvin \((K)\), pressure \((P)\)is in atm.

\(\begin{aligned}{}\frac{P}{{RT}}& = \frac{n}{V}\\\frac{{P \cdot (\mu )}}{{RT}} &= \frac{{n \cdot (\mu )}}{V}\\\frac{{P \cdot (\mu )}}{{RT}} &= \frac{m}{V}\\\frac{{P \cdot (\mu )}}{{RT}} &= \rho {\rm{ }}\\{\rm{Density }}(\rho ) &= \frac{{P \cdot (\mu )}}{{RT}}\end{aligned}\)

where \(\mu = \)molar mass.

Consider the given information.

Volume \((V) = 3L\).

Pressure \((P) = 10{\rm{kPa}}\).

Temperature \((T) = {28^\circ }C\).

Molar mass \((\mu )\)of \({{\bf{O}}_2}\)

Atomic weight of Carbon \((O) = 16\) grams.

So, molecular weight of \({{\rm{O}}_2} = (2) \cdot (16)\) grams.

\( = 32\)grams.

Molar mass of \({{\rm{O}}_2} = 32\frac{{{\rm{ gram }}}}{{{\rm{mol}}}}\).

Converting the temperature in degree Celsius to kelvins.

We have, \({0^\circ } = (0) + 273.15K\)

So,

\(\begin{aligned}{}{28^\circ }C = (28) + 273.15K\\ = 301.15K\end{aligned}\)

First find the density.

\(\begin{aligned}{}{\mathop{\rm Density}\nolimits} (\rho ) &= \frac{{P \cdot (\mu )}}{{RT}}\\ &= \frac{{(10) \cdot (32)}}{{(0.08206) \cdot (301.15)}}\\ &= \frac{{320}}{{24.712}}\\ &= 12.94\frac{{{\mathop{\rm gram}\nolimits} }}{L}\end{aligned}\)

We know that,

\(\begin{aligned}{}\rho = \frac{m}{V}\;\\{\rm{m}} = \rho \cdot V\end{aligned}\)

Where,

\(\begin{aligned}{}\rho &= {\rm{ density }}\\m &= {\rm{ mass }}\\V &= {\rm{ Volume }}\\{\mathop{\rm mass}\nolimits} (m) &= \rho \cdot V\\ &= (12.94) \cdot (3)\\ &= 38.8{\rm{ grams}}\end{aligned}\)

Therefore, mass \(\left( m \right) = 38.8\)grams.