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Chapter 13: Problem 52

In intergalactic space, there is an average of about one hydrogen atom per \(\mathrm{cm}^{3}\) and the temperature is \(3 \mathrm{K}\) What is the absolute pressure?

Short Answer

Expert verified
Answer: The approximate absolute pressure in intergalactic space is 4.14 x 10鈦宦光伕 Pa.

Step by step solution

01

Data given in the question

We are given the following data: - Average density of hydrogen atoms: 1 atom/\(\mathrm{cm}^{3}\) - Temperature: \(3 \mathrm{K}\) We will use the ideal gas law to calculate the absolute pressure.
02

Ideal gas law and constants

The ideal gas law can be expressed as: \(P V = n R T\) where \(P\) - Pressure (in Pascals) \(V\) - Volume (in cubic meters) \(n\) - number of moles of gas \(R\) - Ideal gas constant (\(8.314 \mathrm{J\, mol^{-1} K^{-1}}\)) \(T\) - Temperature (in Kelvin) Also, we need to recall: - Avogadro's number: \(N_A = 6.022 \times 10^{23}\, \mathrm{atoms/mol}\) - Boltzmann's constant: \(k_B = 1.38 \times 10^{-23}\, \mathrm{J/K}\) The pressure we want to find is: \(P = \frac{n R T}{V} = \frac{N k_B T}{V' * V}\) where \(N\) - number of hydrogen atoms per unit volume \(V'\) - Volume of unit hydrogen atom \(V\) - total volume of the space
03

Number of hydrogen atoms per cm鲁

We are given that there is one hydrogen atom per cm鲁. Hence, the number of hydrogen atoms per unit volume is: \(N = \frac{1\, \mathrm{atom}}{\mathrm{cm}^{3}}\).
04

Convert volume to meters鲁

It's important to note that all units in the formula should be in SI (International System) units. So, we need to convert volume from cubic centimeters to cubic meters: \(V' = \frac{1\, \mathrm{cm}^3}{10^6} = 10^{-6}\, \mathrm{m}^3\)
05

Apply the formula for pressure

Now we can use the ideal gas law to calculate the absolute pressure in intergalactic space: \(P = \frac{N k_B T}{V' * V} = \frac{1\, \mathrm{atom/cm^3} \cdot 1.38 \times 10^{-23}\, \mathrm{J/K} \cdot 3 \mathrm{K}}{10^{-6}\, \mathrm{m^3}}\) \(P = 4.14 \times 10^{-18} \, \mathrm{Pa}\) In intergalactic space, the absolute pressure is approximately \(4.14 \times 10^{-18}\, \mathrm{Pa}\).

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