Question

A $1.0\mathrm{m}\times 1.0\mathrm{m}\times 1.0\mathrm{m}$ cube of nitrogen gas is at $20°\mathrm{C}$ and $1.0atm.$ Estimate the number of molecules in the cube with a speed between $700\mathrm{m}/\mathrm{s}$ and $1000\mathrm{m}/\mathrm{s}$.

Short answer

The number of molecules in the cube with a speed between $700\mathrm{m}/\mathrm{s}$ and$1000\mathrm{m}/\mathrm{s}$is$N=2.50\times {10}^{25}$molecules.

Step-by-step solution

Given Information

Nitrogen cube dimension $=1.0\mathrm{m}\times 1.0\mathrm{m}\times 1.0\mathrm{m}$

Temperature $={20}^{\circ }C$

Pressure$=1.0atm$

Explanation 

According to the ideal gas law, the volume of the container $V$, the pressure $p$exerted by the gas, the temperature $T$of the gas, and the number of moles $n$of the gas in the container are all related.

localid="1648276021831" $pV=nRT\to \left(1\right)$

Universal gas constant $=R$and in SI unit its value is,

$R=8.31\mathrm{J}/\mathrm{mol}·\mathrm{K}$

By using Avogadro's number $NA$, we get the number of moles $n$by knowing the number of molecules $N$.

$n=\frac{N}{N_{\mathrm{A}}}$

Use the expression of$n$above equation (1) and solve for $N,$

$pV=nRT$

$pV=\left(\frac{N}{N_{\mathrm{A}}}\right)RT$

localid="1648276044018" $N=\frac{pV{N}_A}{RT}\to \left(2\right)$

Explanation

To convert Celsius to Kelvin, we must first convert the units between the two scales. Kelvins are the units for the Kelvin scale. It can be converted between the two scales by using equation (18.7).

$T_{\mathrm{K}}=T_{\mathrm{C}}+273\to \left(3\right)$

Plug the value for $T_{\mathrm{C}}=20°\mathrm{C}$into equation (3) to get $T_{\mathrm{K}}$

$T_{\mathrm{K}}=T_{\mathrm{C}}+273$

$=20°\mathrm{C}+273$

$=293\mathrm{K}$

Calculate the pressure in $\mathrm{Pa}$,

$p=(1\mathrm{atm})\left(\frac{1.01325\times {10}^5\mathrm{Pa}}{1\mathrm{atm}}\right)=1.01325\times {10}^5\mathrm{Pa}$

Put the values of $p,V,N_A,R$and $T$into equation (2)

$N=\frac{pV{N}_A}{RT}$

$=\frac{\left(1.01325\times {10}^5\mathrm{Pa}\right)\left(1.0{\mathrm{m}}^3\right)\left(6.023\times {10}^{23}\right)}{(8.314\mathrm{J}/\mathrm{mol}\cdot \mathrm{K})\left(293\mathrm{K}\right)}$

$=2.50\times {10}^{25}\text{molecules}$

Final Answer

Hence, the number of molecules in the cube is$N=2.50\times {10}^{25}\text{molecules.}$