Question

What is the density of laughing gas, dinitrogen monoxide, \({{\rm{N}}_2}{\rm{O}}\), at a temperature of \(325\;{\rm{K}}\)and a pressure of \(113.0{\rm{kPa}}\)?

Short answer

The density of laughing gas, dinitrogen monoxide is \(5.254\frac{{{\mathop{\rm gram}\nolimits} }}{L}\).

Step-by-step solution

Definition of volume

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

Define density and \(\mu \)

Let us solve the given problem.

\(PV = nRT\)

where,

\(P = \)pressure of the gas.

\(V = \)volume of the gas.

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

\(T = \)temperature of the gas.

\(\begin{aligned}{}R & = {\rm{ Ideal gas constant}}\\ & = 0.08206L{\rm{ atm mo}}{{\rm{l}}^{ - 1}}{K^{ - 1}}\end{aligned}\)

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

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

Therefore density \((\rho ) = \frac{{P \cdot (\mu )}}{{RT}}\)where \(\mu = \)molar mass.

Converting pressure

Consider the given information.

Pressure\((P) = 113{\rm{kPa}}\).

Temperature\((T) = 325K\).

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

Atomic weight of Carbon\(({\rm{N}}) = 14\)grams.

Atomic weight of Hydrogen\((0) = 16\)gram.

So, molecular weight of\({{\rm{N}}_2}{\rm{O}} = (2) \cdot (14) + (16)\)grams.

\( = 28 + 16\)grams.

\( = 44\)grams.

Molar mass of\({{\rm{N}}_2}{\rm{O}} = 44\frac{{{\rm{gram}}}}{{{\rm{mol}}}}\)

Converting pressure in\({\rm{kPa}}\)into atm.

We have,\(1{\rm{kPa}} = 0.0098\;{\rm{atm}}\).

So,

\(325{\rm{kPa}} = (325) \cdot (0.0098){\rm{atm}}.\)

=\(3.185{\rm{ }}atm.\)

Find density

Calculate density in order to get the final answer.

\(\begin{aligned}{}{\mathop{\rm Density}\nolimits} (\rho ) &= \frac{{P \cdot (\mu )}}{{RT}}\backslash \\ &= \frac{{(3.185) \cdot (44)}}{{(0.08206) \cdot (325)}}\\ &= \frac{{140.14}}{{26.669}}\\ &= 5.254\frac{{{\mathop{\rm gram}\nolimits} }}{L}\end{aligned}\)

Therefore, the calculated density is \(5.254\frac{{{\mathop{\rm gram}\nolimits} }}{L}\).