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Chapter 13: Q84GP (page 359)

In outer space, the density of matter is about one atom per cm3, mainly hydrogen atoms, and the temperature is about 2.7 K. Calculate the RMSspeed of these hydrogen atoms and the pressure (in atmospheres).

Short Answer

Expert verified

The RMS speed of hydrogen atoms is \(2.6 \times {10^2}\;{\rm{m/s}}\), and the pressure is \(3.7 \times {10^{ - 22}}\;{\rm{atm}}\).

Step by step solution

01

Step 1:Understanding of root-mean square (RMS) speed

The square root of the mean of the square of the molecules鈥� speed in a gas at a certain temperature Tis termed as the root-mean square speed of that gas.

The RMS speed is given as

\({v_{{\rm{rms}}}} = \sqrt {\frac{{3RT}}{M}} \).

Here, R is the universal gas constant, T is the temperature, and M is the molecular mass of the gas.

02

Given Data

The temperature of outer space is \(T = 2.7\;{\rm{K}}\).

The molecular mass of hydrogen gas is \(M = 1\;{\rm{g/mol}} = {10^{ - 3}}\;{\rm{kg/mol}}\).

The volume occupied by each atom in the spaceis\(V = 1\;{\rm{c}}{{\rm{m}}^{\rm{3}}} = 1 \times {10^{ - 6}}\;{{\rm{m}}^{\rm{3}}}\).

03

Determination of RMS speed of the hydrogen atoms

The RMS speed of hydrogen atoms is given as

\({v_{{\rm{rms}}}} = \sqrt {\frac{{3RT}}{M}} \).

Substitute the values into the above expression.

\(\begin{aligned}{v_{{\rm{rms}}}} &= \sqrt {\frac{{3 \times 8.314\;{\rm{J/mol - K}} \times 2.7\;{\rm{K}}}}{{{{10}^{ - 3}}\;{\rm{kg/mol}}}}} \\ &= \sqrt {6.7343 \times {{10}^4}} \\ &= 2.6 \times {10^2}\;{\rm{m/s}}\end{aligned}\)

04

Determination of the pressure

According to the ideal gas law,

\(\begin{aligned}PV &= NkT\\P &= \frac{{NkT}}{V}\end{aligned}\).

Here, P is the pressure, V is the volume, N is the number of atoms, and k is the Boltzmann constant.

Substitute the values into the above expression.

\(\begin{aligned}P &= \frac{{\left( 1 \right)\left( {1.38 \times {{10}^{ - 23}}\;{\rm{J/K}}} \right)\left( {2.7\;{\rm{K}}} \right)}}{{\left( {1 \times {{10}^{ - 6}}\;{{\rm{m}}^{\rm{3}}}} \right)}}\\ &= 3.726 \times {10^{ - 17}}\;{\rm{Pa}}\\ &= 3.726 \times {10^{ - 17}}\;{\rm{Pa}} \times \left( {\frac{{1\;{\rm{atm}}}}{{1.013 \times {{10}^5}\;{\rm{Pa}}}}} \right)\\ &= 3.7 \times {10^{ - 22}}\;{\rm{atm}}\end{aligned}\)

Thus, the pressure is \(3.7 \times {10^{ - 22}}\;{\rm{atm}}\).

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