An expensive vacuum system can achieve a pressure as low as 1.00×10−8 N/m2 at 21.5∘C.
The Correct Answer and Explanation is:
Correct Answer: The number of atoms NN in a cubic centimeter at the given pressure and temperature is approximately N≈2.46×106N \approx 2.46 \times 10^6 atoms
Explanation: To determine the number of atoms in a given volume of gas, we use the ideal gas law in the form:
PV=NkTPV = NkT
Here, PP is pressure in pascals (N/m²), VV is volume in cubic meters, NN is the number of atoms, kk is Boltzmann’s constant, and TT is temperature in kelvin.
Given Values:
- Pressure P=1.00×10−8P = 1.00 \times 10^{-8} N/m²
- Temperature T = 21.5\,^\circ\text{C} = 21.5 + 273.15 = 294.65 K
- Boltzmann constant k=1.38×10−23k = 1.38 \times 10^{-23} J/K
- Volume V=1.00 cm3=1.00×10−6 m3V = 1.00\,\text{cm}^3 = 1.00 \times 10^{-6} \,\text{m}^3
Rearranging the ideal gas law to solve for NN:
N=PVkTN = \frac{PV}{kT}
Step-by-step Calculation: Numerator:
(1.00×10−8)×(1.00×10−6)=1.00×10−14(1.00 \times 10^{-8}) \times (1.00 \times 10^{-6}) = 1.00 \times 10^{-14}
Denominator:
(1.38×10−23)×(294.65)=4.06717×10−21(1.38 \times 10^{-23}) \times (294.65) = 4.06717 \times 10^{-21}
Finally:
N=1.00×10−144.06717×10−21≈2.46×106N = \frac{1.00 \times 10^{-14}}{4.06717 \times 10^{-21}} \approx 2.46 \times 10^6
This result shows that even under ultra-low pressure, millions of atoms still occupy one cubic centimeter. It highlights the sensitivity of atomic density to pressure and temperature, reinforcing the utility of the ideal gas equation in approximating microscopic quantities under macroscopic conditions. This is particularly valuable in vacuum physics where precision matters. The use of Boltzmann’s constant bridges the microscopic scale of individual particles with measurable macroscopic quantities like pressure and temperature. This approach is widely used in both theoretical and applied physics.
