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rjkz [21]
3 years ago
11

The average distance between collisions for atoms in a real gas is known as the mean free path (see pp. 298-9 in McKay). Which o

f the answers below best represents the mean free path for gas molecules at STP with atomic radii on the order of 1.0 x 10-10 m? The average distance between collisions for atoms in a real gas is known as the mean free path (see pp. 298-9 in McKay). Which of the answers below best represents the mean free path for gas molecules at STP with atomic radii on the order of 1.0 x 10-10 m?
A. 30 nm
B. 3000 nm
C. 300 nm
D. 3 nm
Physics
1 answer:
dusya [7]3 years ago
4 0

Answer:

300 nm

Explanation:

R = Gas constant = 8.314 J/molK

r = Atomic radii = 1\times 10^{-10}\ m

d = Atomic diameter = 2r=2\times 10^{-10}\ m

At STP

T = Temperature = 273.15 K

P = Pressure = 100 kPa

N_A = Avogadro's number = 6.022\times 10^{23}

The mean free path is given by

\lambda=\frac{RT}{\sqrt2d^2N_AP}\\\Rightarrow \lambda=\frac{8.314\times 273.15}{\sqrt2 \pi \times (2\times 10^{-10})^2\times 6.022\times 10^{23}\times 100000}\\\Rightarrow \lambda=2.12165\times 10^{-7}\ m=212.165\times 10^{-9}\ m=212.165\ nm

The answer that best represents the mean free path for gas molecules is 300 nm

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Nuclear waste disposal is one of the largest issues with nuclear power. Cesium-137 is one of the high level waste products in an
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1. A DC-10 jumbo jet maintains an airspeed of 550 mph in a southwesterly direction. The velocity of the jet stream is a constant
Vladimir79 [104]

Answer:

The magnitude of actual velocity is <u>496.67 mph</u> and its direction is <u>51.54° with the x axis in the third quadrant</u>.

Explanation:

Given:

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Velocity of jet stream from west to east direction (v_s)=80\ mph

First let us draw a vectorial representation of the above velocity vectors.

Consider the south direction as negative y axis and west direction as negative x axis.

From the diagram,

The velocity of the jet can be represented as:

\vec{v_j}=-550\cos(45)\vec{i}+(-550\sin(45)\vec{j} )\\\\\vec{v_j}=-388.91\vec{i}-388.91\vec{j}\ mph

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Now, the vector sum of the above two vectors gives the actual velocity of the aircraft. So, the resultant velocity is given as:

\vec{v}=\vec{v_j}+\vec{v_s}\\\\\vec{v}=-388.91\vec{i}-388.91\vec{j}+80\vec{i}\\\\\vec{v}=(-388.91+80)\vec{i}-388.91\vec{j}\\\\\vec{v}=(-308.91)\vec{i}-388.91\vec{j}

Now, magnitude is given as the square root of sum of the squares of the 'i' and 'j' components. So,

|\vec{v}|=\sqrt{(-308.91)^2+(-388.91)^2}\\\\|\vec{v}|=496.67\ mph

As the horizontal and vertical components of actual velocity negative, the resultant vector makes an angle \theta with the x axis in the third quadrant.

The direction is given as:

\theta=\tan^{-1}(\frac{v_y}{v_x})\\\\\theta=\tan^{-1}(\frac{-388.91}{-308.91})\\\\\theta=51.54\°(Third\ quadrant)

Therefore, the magnitude of actual velocity is 496.67 mph and its direction is 51.54° with the x axis in the third quadrant.

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