An expensive vacuum system can achieve a pressure as low as at . How many atoms are there in a cubic centimeter at this pressure and temperature? The Boltzmann constant . ms
The Correct Answer and Explanation is:
To solve this problem, we will estimate the number of atoms in 1 cubic centimeter (cm³) of gas under extremely low pressure using the ideal gas law. We’ll assume the question refers to a very low pressure such as 1.00 × 10⁻⁹ atm, which is typical for high-vacuum systems, and a room temperature of T = 298 K. The Boltzmann constant is given as:k=1.38×10−23 J/Kk = 1.38 \times 10^{-23} \ \text{J/K}k=1.38×10−23 J/K
Step 1: Use the Ideal Gas Law in molecular form
The molecular form of the ideal gas law is:PV=NkTP V = N k TPV=NkT
Where:
- PPP = pressure in Pascals
- VVV = volume in cubic meters (m³)
- NNN = number of atoms
- kkk = Boltzmann constant
- TTT = temperature in Kelvin
Step 2: Convert units
We are given:
- P=1.00×10−9 atmP = 1.00 \times 10^{-9} \ \text{atm}P=1.00×10−9 atm
- Convert atm to Pascals: 1 atm=1.013×105 Pa1 \ \text{atm} = 1.013 \times 10^5 \ \text{Pa}1 atm=1.013×105 Pa
So:P=1.00×10−9×1.013×105=1.013×10−4 PaP = 1.00 \times 10^{-9} \times 1.013 \times 10^5 = 1.013 \times 10^{-4} \ \text{Pa}P=1.00×10−9×1.013×105=1.013×10−4 Pa
Also:
- V=1 cm3=1.00×10−6 m3V = 1 \ \text{cm}^3 = 1.00 \times 10^{-6} \ \text{m}^3V=1 cm3=1.00×10−6 m3
Step 3: Solve for NNN
N=PVkT=(1.013×10−4)(1.00×10−6)(1.38×10−23)(298)N = \frac{P V}{k T} = \frac{(1.013 \times 10^{-4})(1.00 \times 10^{-6})}{(1.38 \times 10^{-23})(298)}N=kTPV=(1.38×10−23)(298)(1.013×10−4)(1.00×10−6)N=1.013×10−104.1124×10−21≈2.46×1010 atomsN = \frac{1.013 \times 10^{-10}}{4.1124 \times 10^{-21}} \approx 2.46 \times 10^{10} \ \text{atoms}N=4.1124×10−211.013×10−10≈2.46×1010 atoms
Final Answer:
There are approximately 2.46 × 10¹⁰ atoms in 1 cubic centimeter at this low pressure and room temperature.
Explanation
In high vacuum systems, we often describe the environment in terms of pressure and temperature to determine how many gas particles remain in a volume. To calculate the number of atoms in a cubic centimeter of gas under low pressure, we apply the ideal gas law in terms of number of particles.
The Boltzmann constant links the macroscopic properties of gases (like pressure and volume) to microscopic behavior (individual atoms or molecules). By rearranging the ideal gas equation PV=NkTPV = NkTPV=NkT, we can isolate NNN, the number of atoms or molecules in the system.
First, the pressure is converted to Pascals since SI units are necessary for this equation. One atmosphere is equal to 101325 Pascals. A high vacuum might reach 1.00 × 10⁻⁹ atm, which converts to 1.013 × 10⁻⁴ Pascals. The volume is also converted to cubic meters: 1 cubic centimeter is 1.00 × 10⁻⁶ cubic meters.
By inserting these values into the equation N=PVkTN = \frac{PV}{kT}N=kTPV, and using the Boltzmann constant k=1.38×10−23 J/Kk = 1.38 \times 10^{-23} \ \text{J/K}k=1.38×10−23 J/K and room temperature 298 K, we calculate the number of atoms to be approximately 2.46 × 10¹⁰.
This result tells us that even at extremely low pressures, there are still billions of atoms in every cubic centimeter, illustrating how small atoms are and how challenging it is to create a true vacuum.
