Find the rms speed of a helium molecule at 321 K
This question has been posted and answered elsewhere.
The answer is sqrt(3RT)
Do the numbers.
To find the root mean square (rms) speed of a helium molecule at a given temperature, we can use the formula:
v = √((3*k*T) / (m))
where:
- v is the rms speed (in meters per second)
- k is the Boltzmann constant (1.38 x 10^-23 J/K)
- T is the temperature in Kelvin
- m is the mass of the helium molecule (4.0026 atomic mass units or amu)
First, we need to convert the temperature from Celsius to Kelvin. The conversion formula is:
T(K) = T(°C) + 273.15
Let's apply this to your question:
T(K) = 321 + 273.15
T(K) = 594.15 K
Now, we can substitute the values into the formula:
v = √((3 * 1.38 x 10^-23 J/K * 594.15 K) / (4.0026 amu))
To proceed further, we need to convert the mass from atomic mass units (amu) to kilograms. We know that:
1 amu = 1.66 x 10^-27 kg
So, the mass of the helium molecule in kilograms will be:
m(kg) = 4.0026 amu * (1.66 x 10^-27 kg / 1 amu)
Now, we can substitute the values and calculate the rms speed:
v = √((3 * 1.38 x 10^-23 J/K * 594.15 K) / (4.0026 amu * 1.66 x 10^-27 kg / 1 amu))
After performing the calculations, the resulting value will give you the rms speed of a helium molecule at 321 K.