At what temperature (in Kelvin) will the average speed of hydrogen mol...
If
are the average speeds of hydrogen and nitrogen molecules at temperatures T
1K and T
2K respectively, we have
...(i)
...(ii)
where m
1 and m
2 are the masses of hydrogen and nitrogen molecules respectively.
Dividing eq. (1) and (2), we get
...(iii)
Given when
and
Substituting these values in (3), we get
or
The most probable speed of nitrogen at temperature T is given by
Here k = 1.38 * 10
-23 joule per degree, T = 22 K
and m = 28 a.m.u. =
∴
= 1.14 * 10
2m/s.
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At what temperature (in Kelvin) will the average speed of hydrogen mol...
To find the temperature at which the average speed of hydrogen molecules is the same as that of nitrogen molecules, we can use the equation for average speed:
v = sqrt((3*k*T)/m)
where v is the average speed, k is the Boltzmann constant (1.38 x 10^-23 J/K), T is the temperature in Kelvin, and m is the molecular mass.
For nitrogen, m = 28 atomic mass units = 28 * 1.66 x 10^-27 kg.
For hydrogen, m = 2 atomic mass units = 2 * 1.66 x 10^-27 kg.
Let's set the average speeds of the two molecules equal to each other:
sqrt((3*k*T_nitrogen)/m_nitrogen) = sqrt((3*k*T_hydrogen)/m_hydrogen)
Simplifying, we can cancel out the square root and the Boltzmann constant:
sqrt(T_nitrogen/m_nitrogen) = sqrt(T_hydrogen/m_hydrogen)
Squaring both sides, we get:
T_nitrogen/m_nitrogen = T_hydrogen/m_hydrogen
Plugging in the values for m_nitrogen, m_hydrogen, and T_nitrogen = 35 + 273.15 K, we can solve for T_hydrogen:
(35 + 273.15) / 28 = T_hydrogen / 2
308.15 / 28 = T_hydrogen / 2
10.72 = T_hydrogen / 2
T_hydrogen = 10.72 * 2
T_hydrogen = 21.44 K
Therefore, the temperature at which the average speed of hydrogen molecules is the same as that of nitrogen molecules kept at 35°C is approximately 21.44 Kelvin.