## College Physics (4th Edition)

(a) $\rho = 1.34~kg/m^3$ (b) $\rho = 1.16~kg/m^3$
(a) We can convert the average molecular mass to units of $kg$: $m = (29~u)(1.66\times 10^{-27}~kg/u) = 4.814\times 10^{-26}~kg$ We can find an expression for the number density: $P~V = N~k~T$ $\frac{N}{V} = \frac{P}{k~T}$ We can find an expression for the mass density: $\rho = \frac{m~N}{V}$ $\rho = \frac{m~P}{k~T}$ We can find the mass density when the temperature is $-10^{\circ}C$ which is $263~K$: $\rho = \frac{m~P}{k~T}$ $\rho = \frac{(4.814\times 10^{-26}~kg)(1.01\times 10^5~Pa)}{(1.38\times 10^{-23}~J/K)(263~K)}$ $\rho = 1.34~kg/m^3$ (b) We can find the mass density when the temperature is $30^{\circ}C$ which is $303~K$: $\rho = \frac{m~P}{k~T}$ $\rho = \frac{(4.814\times 10^{-26}~kg)(1.01\times 10^5~Pa)}{(1.38\times 10^{-23}~J/K)(303~K)}$ $\rho = 1.16~kg/m^3$