Question

03. The lowest pressure (the best vaccum) that can be created in laboratory at \( 27^{\circ} C \) is \( 10^{-11} mm \) of Hg .At this pressure, the number of ideal gas molecules per \( cm ^{3} \) will be :- 1) \( 3.22 \times 10^{12} \) 2) \( 1.61 \times 10^{12} \) 3) \( 3.21 \times 10^{6} \) 4) \( 3.22 \times 10^{5} \)

Comments

To calculate the number of ideal gas molecules per cubic centimeter, we can use the ideal gas equation:

PV = nRT

Where:

• P is the pressure

• V is the volume

• n is the number of moles

• R is the universal gas constant (8.314 J/mol^.K)

• T is the temperature in Kelvin

Given the temperature T = 27°C = 300 K and the pressure P = 10⁻¹¹ mm Hg:

Convert the pressure to Pascals (Pa):

\(10^{-11} \, \text{mm Hg} \times 133.322 \, \text{Pa/mm Hg} = 1.33322 \times 10^{-9} \, \text{Pa}\)

Rearrange the ideal gas equation to solve for n/V (number of moles per unit volume):

n/V = PRT

Substitute the values:

\(\frac{1.33322 \times 10^{-9}}{8.314 \times 300} = 5.35 \times 10^{-13} \, \text{mol/m}^3\)

Convert  \(\text{mol/m}^3 \ to \text{ mol/cm}^3:\)

5.35×10⁻¹³ \(\text{mol/m}^3\) = 5.35×10⁻¹⁹  \(\text{mol/cm}^3\)

Using Avogadro's number (\(6.022 \times 10^{23} \, \text{molecules/mol}\)):

\(5.35 \times 10^{-19} \times 6.022 \times 10^{23} = 3.22 \times 10^5 \, \text{molecules/cm}^3\)

So, the number of ideal gas molecules per cubic centimeter is:

\(\boxed{3.22 \times 10^5}\)

The correct answer is option 4.