91Ó°ÊÓ

The molar mass of achemical compoundis defined as themassof a sample of that compound divided by theamount of substancein that sample, measured inmoles.

Consider the given data as below.

Molar mass of ${\mathrm{N}}_2$, $M=28×{10}^{-3}kg$

And 94.7% of its molecules move with a given speed (root mean square speed). And this given fraction is 0.974 represent the molecules in an ideal gas with speeds less than various.

$\frac{v}{v_{rms}}=1.60$

Before calculating the temperature of the 94 - 7% of $N_2$ molecules you must know that the $N_2$molecules have a root mean square speed 11m less than 1 where v is the molecular speed and the ratio between them is given by

$\frac{v}{v_{rms}}=1.60$

Therefore,

$v_{rms}=\frac{v}{1.690}$ ….. (1)

By apply the root mean square speed equation which employ the relation between root mean square speed 11 m , molar

By apply the root mean square speed equation which employ the relation between root mean square speed 0 m , molar

$v_{rms}=\sqrt{\frac{3RT}{M}}$ ….. (2)

Here, M is the mass, R is the gas constant and T is the temperature.

And Compare equation (1) and (2) as below.

$\frac{v}{1.60}=\sqrt{\frac{3RT}{M}}$

Now take the square for both sides and solve for T and employ equation (2) by

$T=\frac{M{v}^2}{3{\left(1.6\right)}^{2}R}$ ….. (3)

Now define the equation that make able to calculate T by known 0 in part (a), (b) and (c).

Substitute $28×{10}^{-3}\raisebox{1ex}{$kg$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.$ for M , $1500\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$ for v , and $8.314J/mol·K$ for R in the above equation.

$\begin{array}{r}T=\frac{28×{10}^{−3}×(1500{)}^2}{3(1.6{)}^2×8.314}\\ =987\mathrm{K}\end{array}$

Therefore, the temperature for 1500 m/s is 987 K.

Substitute $28×{10}^{-3}\raisebox{1ex}{$kg$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.$ for M , $1000\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$ for v , and 8.314 J/mol.K for R into equation (3) as below.

$\begin{array}{r}T=\frac{{\mathrm{Mv}}^2}{3(1.6{)}^2\mathrm{R}}\\ =\frac{28×{10}^{−3}×(1000{)}^2}{3(1.6{)}^2×8.314}\\ =438\mathrm{K}\end{array}$

Therefore, the temperature at 1000 m/s is 438 K .

Substitute $28×{10}^{-3}\raisebox{1ex}{$kg$}\!\left/ \!\raisebox{-1ex}{$mol$}\right.$ for M , $500\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$ for v , and 8.314 J/mol.K for R into equation (3) as below.

$\begin{array}{r}T=\frac{M{v}^2}{3(1.6{)}^{2}R}\\ =\frac{28×{10}^{−3}×(500{)}^2}{3(1.6{)}^2×8.314}\\ =110\mathrm{K}\end{array}$

Therefore, the temperature at 500 m/s is 110 K .