Question

The semiconductor industry manufactures integrated circuits in large vacuum chambers where the pressure is 1.0 * 10-10 mm of Hg.

a. What fraction is this of atmospheric pressure?

b. At T = 20$°$C, how many molecules are in a cylindrical chamber 40 cm in diameter and 30 cm tall?

Short answer

The answers of (a) is $\frac{Pvacuum}{P}=1.31\times {10}^{-13}$

(b) is$N=1.22\times {10}^{11}molecules$

Step-by-step solution

Description 

The semiconductor industry manufactures integrated circuits in large vacuum chambers where the pressure is 1.0 * 10-10 mm of Hg.

The atmospheric pressure is $\$p=\$1atm$

The vacuum chamber's pressure is$Pvacuum=1\times 1^{-10}mmHg.$

Result of finding the fraction and molecules 

The atmospheric pressure is $\$p=\$1atm$

The conversion of unit from atm to mmHg through

$p=\left(1atm\right)\left(\frac{760mmHg}{1atm}\right)=760mmHg$

Ratio is the fraction of pressure's chamber and the atmospheric pressure.

Thus the calculation is:

$$\begin{gathered} f=\frac{Pvacuum}{p} \\ =\frac{1.0\times {10}^{-10}mmHg}{70mmHg} \\ =1.31\times {10}^{-13} \end{gathered}$$

The radius of cylinder is:

$r=\frac{d}{2}=\frac{40cm}{2}=20cm=0.20m$

The volume of cylinder is:

$\begin{array}{rcl}V& =& {\mathrm{\pi r}}^2\mathrm{L}\\ & =& \mathrm{\pi }{(0.2\mathrm{m})}^2(0.30\mathrm{m})\\ & =& 0.038{\mathrm{m}}^3\end{array}$

The ideal law of gas has provided to the form of equation (18.13) is:

$pV=nRT$ (1)

In this canre R is universal gas contact and its value on SI unit is:

$R=8.31LJ/mol.K$

Another law that is effectively related to the number of molecules N of gas than the number of moles is in the form of equation (18.16) is:

$pV=\frac{N}{N_A}RT$ (2)

In this case, the Avogadro's number is$N_A$. The equation (2)on the basis of N is:
$N=\frac{p{N}_{A}V}{RT}$ (3)

The pascal is calculated through

$p=(1.0\times {10}^{-10}mmHg)\left(\frac{133.322Pa}{1mmHg}\right)=1.3\times {10}^{-8}Pa$

The conversion between Kelvin scale and Celcius scale is provided in the form of equation (18.7):

$T_K=T_C=273$ (4)

Thus, the value of $T_C=20°C$into equation (3) in order to get $T_K$

$$\begin{gathered} T_K=T_c+273 \\ =20°C+273 \\ =293K \end{gathered}$$

Thus, plug the values for$T,R,V,N_{A,p}$ into equation (3) in order to get N:
$$\begin{gathered} N=\frac{p{N}_{A}V}{RT} \\ =\frac{(1.3\times {10}^{-8}Pa)(6.02\times {10}^{23})(0.038m^3)}{(8.31J/mol.K)\left(293K\right)} \\ =1.22\times {10}^{11}molecules \end{gathered}$$

Therefore, The answers of (a) is $\frac{Pvacuum}{P}=1.31\times {10}^{-13}$as well as (b) is$N=1.22\times {10}^{11}molecules$.