Compute the rms speed of a nitrogen molecule at 31.0°C. At what temperature will the rms speed be half that value?
At what temperature will the rms speed be twice that value?
T = 31 C = 304 K
Mass M = 28 g/(6.02*10^23)
= 4.65*10^-23 g = 4.65*10^-26 kg
The average kinetic energy per molecule is
(1/2) M V^2 = (3/2) k T
where k = 1.38*10^-23 Joule/K is Boltzmann's constant and V^2 is the average value of (molecular velcoity)^2
(For explanation, see http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/kintem.html )
Solve for V^2, making sure T is in K and M is in kg. The V^2 that you get will be in m^2/s^2.
The square root of V^2 is the rms (root mean square) velocity.
Thanks!
First we need to calculate V^2:
(1/2) M V^2 = (3/2) k T
V^2 = (3 k T) / M
V^2 = (3 * 1.38*10^-23 Joule/K * 304 K) / 4.65*10^-26 kg
V^2 ≈ 2845139.78 m^2/s^2
Now we can find the rms speed by taking the square root of V^2:
Vrms = √(2845139.78 m^2/s^2) ≈ 1686.75 m/s
To find the temperature at which the rms speed is half of this value, we can use this equation:
V^2 = (3 k T) / M
Or we can use the relationship between T and Vrms:
V^2 = (3 k T1) / M = (1/4) * (3 k T2) / M
T2 = (1/4) * T1
T2 = (1/4) * 304 K ≈ 76 K
To find the temperature at which the rms speed is twice the original value, we can use the same equation:
V^2 = (3 k T) / M
Or we can use the relationship between T and Vrms:
V^2 = (3 k T1) / M = (4) * (3 k T3) / M
T3 = (4) * T1
T3 = (4) * 304 K ≈ 1216 K
So the rms speed will be half the original value at 76 K, and twice the original value at 1216 K.
To compute the rms speed of a nitrogen molecule, we can use the equation:
(1/2) M V^2 = (3/2) k T
where M is the mass of the molecule, V is the average velocity of the molecule, k is Boltzmann's constant, and T is the temperature in Kelvin.
Given:
T = 31°C = 304 K
M = 4.65*10^-26 kg
k = 1.38*10^-23 J/K
We need to solve the equation for V^2. Rearranging the equation, we have:
V^2 = (3 k T) / M
Plugging in the values, we have:
V^2 = (3 * 1.38*10^-23 J/K * 304 K) / 4.65*10^-26 kg
Calculating this expression, we get:
V^2 = 5.68 * 10^2 m^2/s^2
Finally, we take the square root of V^2 to find the rms velocity:
V = sqrt(5.68 * 10^2) m/s
= 23.85 m/s (approximately)
Now, to find the temperature at which the rms speed will be half the value, we can set up a proportion:
Using the equation V^2 = (3 k T) / M, we can write:
(V_h/2)^2 = (3 k T_h) / M
where V_h/2 is the rms speed we want to find, and T_h is the corresponding temperature.
Dividing the equation by 4, we get:
(V_h^2) / 4 = (3 k T_h) / M
Rearranging the equation, we have:
T_h = (V_h^2 * M) / (12 k)
Substituting V_h/2 = (1/2) * 23.85 m/s and the known values for M and k, we can solve for T_h:
T_h = ((1/2) * 23.85)^2 * 4.65*10^-26 kg / (12 * 1.38*10^-23 J/K)
Calculating this expression, we find:
T_h ≈ 6.33 K
Therefore, the temperature at which the rms speed will be half the original value is approximately 6.33 K.
To find the temperature at which the rms speed will be twice the original value, we can set up a similar proportion:
(V_2)^2 = (3 k T_2) / M
where V_2 is the desired rms speed, and T_2 is the corresponding temperature.
Dividing the equation by 2, we get:
(V_2^2) / 2 = (3 k T_2) / M
Rearranging the equation, we have:
T_2 = (V_2^2 * M) / (6 k)
Substituting V_2 = 2 * 23.85 m/s and the known values for M and k, we can solve for T_2:
T_2 = (2 * 23.85)^2 * 4.65*10^-26 kg / (6 * 1.38*10^-23 J/K)
Calculating this expression, we find:
T_2 ≈ 96.8 K
Therefore, the temperature at which the rms speed will be twice the original value is approximately 96.8 K.
To compute the rms speed of a nitrogen molecule at 31.0°C (304 K), we can use the equation:
(1/2) M V^2 = (3/2) k T
where M is the mass of the molecule, V^2 is the average value of the square of the molecular velocity, k is Boltzmann's constant (1.38 * 10^-23 J/K), and T is the temperature in Kelvin.
First, let's calculate V^2:
V^2 = (3/2) k T / M
Plugging in the values:
M = 4.65 * 10^-26 kg
k = 1.38 * 10^-23 J/K
T = 304 K
V^2 = (3/2) * (1.38 * 10^-23 J/K) * (304 K) / (4.65 * 10^-26 kg)
V^2 ≈ 6.661 * 10^3 m^2/s^2
Taking the square root of V^2 gives us the rms velocity:
V_rms = √(6.661 * 10^3 m^2/s^2)
V_rms ≈ 81.58 m/s
To calculate the temperature at which the rms speed will be half the value, we can use the same equation:
V_rms = √[(3/2) * k * T / M]
We can solve for T by rearranging the equation:
T = (2 * M * (V_rms^2)) / (3 * k)
Plugging in the values:
V_rms/2 = 0.5 * 81.58 m/s = 40.79 m/s
M = 4.65 * 10^-26 kg
k = 1.38 * 10^-23 J/K
T = (2 * (4.65 * 10^-26 kg) * (40.79 m/s)^2) / (3 * (1.38 * 10^-23 J/K))
T ≈ 49.23 K
Thus, the temperature at which the rms speed will be half the value is approximately 49.23 K.
To calculate the temperature at which the rms speed will be twice the value, we can use the same equation:
V_rms = √[(3/2) * k * T / M]
We can solve for T by rearranging the equation:
T = (2 * M * (V_rms^2)) / (3 * k)
Plugging in the values:
2 * V_rms = 2 * 81.58 m/s = 163.16 m/s
M = 4.65 * 10^-26 kg
k = 1.38 * 10^-23 J/K
T = (2 * (4.65 * 10^-26 kg) * (163.16 m/s)^2) / (3 * (1.38 * 10^-23 J/K))
T ≈ 978.47 K
Thus, the temperature at which the rms speed will be twice the value is approximately 978.47 K.