Question 20.5: Molecular Speeds in a Hydrogen Gas A 0.500-mol sample of hyd...
Molecular Speeds in a Hydrogen Gas
A 0.500-mol sample of hydrogen gas is at 300 K.
(A) Find the average speed, the rms speed, and the most probable speed of the H_2 molecules.
(B) Find the number of molecules with speeds between 400 m/s and 401 m/s.
(A) Conceptualize Imagine the huge number of particles in a macroscopic sample of gas, all moving in random directions with different speeds.
Categorize We are dealing with a very large number of particles, so we can use the Maxwell–Boltzmann speed distribution function.
Analyze Use Equation 20.44 to find the average speed:
v_{\text {avg }}=\sqrt{\frac{8 k_{B} T}{\pi m_0}}=1.60 \sqrt{\frac{k_{B} T}{m_0}} (20.44)
\begin{aligned}v_{\text {avg }} & =1.60 \sqrt{\frac{k_{B} T}{m_0}}=1.60 \sqrt{\frac{\left(1.38 \times 10^{-23} J /K\right)(300 K)}{2\left(1.67 \times 10^{-27} kg\right)}} \\& =1.78 \times 10^3 m/s\end{aligned}Use Equation 20.43 to find the rms speed:
v_{\text {rms }}=\sqrt{\overline{v^2}}=\sqrt{\frac{3 k_{B} T}{m_0}}=1.73 \sqrt{\frac{k_{B} T}{m_0}} (20.43)
\begin{aligned}v_{\text{rms}} & =1.73 \sqrt{\frac{k_{B} T}{m_0}}=1.73 \sqrt{\frac{\left(1.38 \times 10^{-23} J/K\right)(300 K)}{2\left(1.67 \times 10^{-27} kg\right)}} \\& =1.93 \times 10^3 m/s\end{aligned}Use Equation 20.45 to find the most probable speed:
v_{mp}=\sqrt{\frac{2 k_{B} T}{m_0}}=1.41 \sqrt{\frac{k_{B} T}{m_0}} (20.45)
\begin{aligned}v_{mp} & =1.41 \sqrt{\frac{k_{B} T}{m_0}}=1.41 \sqrt{\frac{\left(1.38 \times 10^{-23} J/K\right)(300 K)}{2\left(1.67 \times 10^{-27} kg\right)}} \\& =1.57 \times 10^3 m/s\end{aligned}(B) Use Equation 20.42 to evaluate the number of molecules in a narrow speed range between v and v + dv:
N_v=4 \pi N\left(\frac{m_0}{2 \pi k_{B} T}\right)^{3 / 2} v^2 e^{-m_0 v^2 / 2 k_{B} T} (20.42)
(1) N_v d v=4 \pi N\left(\frac{m_0}{2 \pi k_{B} T}\right)^{3 / 2} v^2 e^{-m_0 v^2 / 2 k_{B} T} d v
Evaluate the constant in front of v²:
\begin{aligned}4 \pi N\left(\frac{m_0}{2 \pi k_{B} T}\right)^{3 / 2}&=4 \pi n N_{A}\left(\frac{m_0}{2 \pi k_{B} T}\right)^{3 / 2}\\& =4 \pi(0.500 \text{ mol})\left(6.02 \times 10^{23} \text{ mol}^{-1}\right)\left[\frac{2\left(1.67 \times 10^{-27} kg\right)}{2 \pi\left(1.38 \times 10^{-23} J/K\right)(300 K)}\right]^{3 / 2} \\& =1.74 \times 10^{14} s^3 /m^3\end{aligned}Evaluate the exponent of e that appears in Equation (1):
-\frac{m_0 v^2}{2 k_{B} T}=-\frac{2\left(1.67 \times 10^{-27} kg\right)(400 m/s)^2}{2\left(1.38 \times 10^{-23} J/K\right)(300 K)}=-0.064 5Evaluate N_v d v using these values in Equation (1) and evaluating v and dv:
\begin{aligned} N_v d v&=\left(1.74 \times 10^{14} s^3 /m^3\right)(400 m/s)^2 e^{-0.064 5}(1 m/s)\\&=2.61 \times 10^{19} \text { molecules }\end{aligned}Finalize In this evaluation, we could calculate the result without integration because dv = 1 m/s is much smaller than v = 400 m/s. Had we sought the number of particles between, say, 400 m/s and 500 m/s, we would need to integrate Equation (1) between these speed limits.