Question

The graph shows hypothetical speed distribution for a sample of 𝑁 gas particles (for $V_0,\frac{dN}{dV}=0$)

• A. The value of $a{V}_0$ is $2N$
• B. The ratio $\frac{V_{avg}}{V_0}$ is equal to $\frac{2}{3}$
• C. The ratio $\frac{V_{rms}}{V_0}$ is equal to $\frac{1}{\sqrt{2}}$
• D. Three-fourths of the total particles have a speed between $0.5V_0$ and $V_0$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A The value of $a{V}_0$ is $2N$
B The ratio $\frac{V_{avg}}{V_0}$ is equal to $\frac{2}{3}$
C The ratio $\frac{V_{rms}}{V_0}$ is equal to $\frac{1}{\sqrt{2}}$
D Three-fourths of the total particles have a speed between $0.5V_0$ and $V_0$


a)

$\frac{dN}{dV}\to v\frac{dN}{dv}=\frac{a}{v_0}\cdot v$

$a{v}_0=2$ (Area under the curve)

Area= $\int ydx=\int \frac{dN}{dv}.dv=\int dN=N$

b)

$\frac{dN}{dv}=\frac{av}{v_0}$

$dN\frac{av}{v_0}dv$

$v_{avg}=\frac{\int dNv}{\int dN}$

$v_{avg}=\frac{{\int }_0^{v_0}\frac{av}{v_0}dvv}{v_0\int \frac{av}{v_0}dv}$

$v_{avg}=\frac{\frac{a}{v_0}{\int }_0^{v_0}{v}^{2}dv}{\frac{a}{v_0}{v}_0^{v_0}vdv}=\frac{2V_0}{3}$

c)

$v_{rms}=\sqrt{meansq}$

$meansq=\frac{\int dN{v}^2}{\int dN}$

$meansq=\frac{{\int }_0^{V_0}\frac{av}{v_0}dv{v}^2}{{\int }_0^{V_0}\frac{av}{v_0}dv}$

$\frac{{\int }_0^{V_0}{v}^{3}dv}{{\int }_0^{V_0}vdv}=\frac{v_0^4}{4\frac{V_0^2}{4}}=0.5V_0^2$

$\begin{array}{cc}{v}_{rms}& =\sqrt{\frac{1}{2}{v}_0^2}\\ V_{rms}& =\frac{v_0}{\sqrt{2}}\\ \frac{V_{rms}}{v_0}& =\frac{1}{\sqrt{2}}\end{array}$


$\frac{v_0}{2}+v_0\to \frac{3}{4}$Number of molequles.