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2 years ago

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What is the root mean square speed of an As4 particle as it is sublimed? (Assume at high temperatures arsenic acts like an ideal

gas; the Boltzmann constant, kB, can be approximated as 1.4 x 10-23 J∙K-1) and the kinetic energy of the gas is equal to 3/2kB T

1 answer:

Norma-Jean [14]2 years ago

5 0

Answer: $v_{rms} =$ 273m/s

Explanation: <u>Root</u> <u>Mean</u> <u>Square</u> <u>Speed</u> of an atom or molecule is the speed of a particle in a gas. It is the average speed a particle in a gas can have.

It can be calculated:

$\frac{1}{2}mv^{2} = \frac{3}{2} k_{B}T$

$v^{2} = \frac{3k_{B}T}{m}$

$v = \sqrt{\frac{3k_{B}T}{m}}$

m is mass of one atom or molecule in kg.

An atom of Arsenic sublimes at 614°C. Converting to Kelvin:

T = 614 + 273 = 887K

Molecular mass of As4 is approximately 0.3kg.

$m = \frac{0.3}{6.10^{23}}$

$m=5.10^{-25}$ kg

Calculating Root mean square speed :

$v = \sqrt{\frac{3*1.4.10^{-23}*887}{5.10^{-25}}$

$v = \sqrt{745.08.10^{2}$

v = 273m/s

<u>The root mean square speed of As4 is approximately </u><u>273m/s</u><u>.</u>

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Two 5.0 mm × 5.0 mm electrodes are held 0.10 mm apart and are attached to 7.5 V battery. Without disconnecting the battery, a 0.

Musya8 [376]

Answer:

A) V = 7.5 V

B) E = 75,000 V/m

C) Q = 16.6 pC

D) V = 7.5 V

E) E = 24,000 V/m

F) Q = 52 pC

Explanation:

Given:

- The Area of plate A = ( 5 x 5 ) mm^2

- The distance between plates d = 0.10 mm

- The thickness of Mylar added t = 0.10 mm

- Voltage supplied by battery V = 7.5 V

Solution:

A) What is the capacitor's potential difference before the Mylar is inserted?

- The potential difference across the two plates is equal to the voltage provided by the battery V = 7.5 V which remains constant throughout.

B) What is the capacitor's electric field before the Mylar is inserted?

- The Electric Field E between the capacitor plates is given by:

E = V / k*d

k = 1 (air) E = 7.5 / 0.10*10^-3

E = 75,000 V/m

C) What is the capacitor's charge Q before the Mylar is inserted?

C = k*A*ε / d

k = 1 (air) C = ( 0.005^2 * 8.85*10^-12 ) / 0.0001

C = 2.213 pF

Q = C*V

Q = 7.5*(2.213)

Q = 16.6 pC

D) What is the capacitor's potential difference after the Mylar is inserted?

- The potential difference across the two plates is equal to the voltage provided by the battery V = 7.5 V which remains constant throughout.

E) What is the capacitor's electric field after the Mylar is inserted?

- The Electric Field E between the capacitor plates is given by:

E = V / k*d

k = 3.13 E = 7.5 / (3.13)0.10*10^-3

E = 24,000 V/m

F) What is the capacitor's charge after the Mylar is inserted?

C = k*A*ε / d

k = 3.13 C = 3.13*( 0.005^2 * 8.85*10^-12 ) / 0.0001

C = 6.927 pF

Q = C*V

Q = 7.5*(6.927)

Q = 52 pC

6 0

2 years ago

Tori experiments with pulleys in physics class. She applies 70 newtons of force to a single pulley to lift a bowling ball. By ad

e-lub [12.9K]

The minimum input force she'll need to lift the ball is 35 N.

Explanation:

Mechanical advantage of a single pulley is 1. As, she applies 70 N of force to lift the bowling ball, so the output force(weight of the ball) is also 70 N.

Now, adding another pulley gives a mechanical advantage of 2. We have,

M.A = (Output Force)/(Input Force)

Substituting the values we get,

$2=\frac{70}{F_{i}}$

$F_{i}=\frac{70}{2}$ = 35 N

Input force equals to 35 N needs to be applied.

6 0

2 years ago

Read 2 more answers

The magnitude J(r) of the current density in a certain cylindrical wire is given as a function of radial distance from the cente

kipiarov [429]

Answer:

$I=68.31\times 10^{-6}\ A$

Explanation:

Given that

J(r) = Br

We know that area of small element

dA = 2 π dr

I = J A

dI = J dA

Now by putting the values

dI = B r . 2 π dr

dI= 2π Br² dr

Now by integrating above equation

$\int_{0}^{I}dI= \int_{r_1}^{r_2}2\pi Br^2 dr$

$I={2\pi B}\times \dfrac{r_2^3-r_1^3}{3}$

Given that

B= 2.35 x 10⁵ A/m³

r₁ = 2 mm

r₂ = 2+ 0.0115 mm

r₂ = 2.0115 mm

$I={2\pi B}\times \dfrac{r_2^3-r_1^3}{3}$

By putting the values

$I={2\pi \times 2.35 \times 10^5 }\times \dfrac{(2.0115\times 10^{-3})^3-(2\times 10^{-3})^3}{3}\ A$

$I=68.31\times 10^{-6}\ A$

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2 years ago

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In the 25-ft Space Simulator facility at NASA's Jet Propulsion Laboratory, a bank of overhead arc lamps can produce light of int

Ugo [173]

Answer:

a. 8.33 x 10 ⁻⁶ Pa

b. 8.19 x 10 ⁻¹¹ atm

c. 1.65 x 10 ⁻¹⁰ atm

d. 2.778 x 10 ⁻¹⁴ kg / m²

Explanation:

Given:

a.

I = 2500 W / m² , us = 3.0 x 10 ⁸ m /s

P rad = I / us

P rad = 2500 W / m² / 3.0 x 10 ⁸ m/s

P rad = 8.33 x 10 ⁻⁶ Pa

b.

P rad = 8.33 x 10 ⁻⁶ Pa *[ 9.8 x 10 ⁻⁶ atm / 1 Pa ]

P rad = 8.19 x 10 ⁻¹¹ atm

c.

P rad = 2 * I / us = ( 2 * 2500 w / m²) / [ 3.0 x 10 ⁸ m /s ]

P rad = 1.67 x 10 ⁻⁵ Pa

P₁ = 1.013 x 10 ⁵ Pa /atm

P rad = 1.67 x 10 ⁻⁵ Pa / 1.013 x 10 ⁵ Pa /atm = 1.65 x 10 ⁻¹⁰ atm

d.

P rad = I / us

ΔP / Δt = I / C² = [ 2500 w / m² ] / ( 3.0 x 10 ⁸ m/s)²

ΔP / Δt = 2.778 x 10 ⁻¹⁴ kg / m²

3 0

2 years ago

29. 2072 Set C Q.No. 10c

Annette [7]

Answer:

$90.2^{\circ}C$

Explanation:

Considering the thermal conductivity of aluminium and brass as $k_{al}=205 W/mK$ and $k_{br}=109 W/mk$ respectively

The temperature at the end of aluminium and brass are given as $T_{al}=150^{\circ}C$ and $T_{br}=20^{\circ}C$ respectively with length of rod L=1.3 m , Length of aluminium $L_{al}=0.8 m$ , length of brass $L_{br}=0.5 m$ and letting temperature at steady state be T

At steady state, thermal conductivity of both aluminium and brass are same hence

$H_{br}=H_{al}$

$k_{al}A\frac {T_H-T}{L_{al}}= k_{br}A\frac {T-T_H}{L_{br}}$

Upon re-arranging

$T=\frac {k_{al}L_{al}T_{br}+k_{al}L_{br}T_{al}}{k_{br}L_{al}+k_{al}L_{br}}$

$(205)\frac {150-T}{0.8}=109\frac {T-20}{0.5}$

$T=\frac {(109*0.8*20)+(205*0.5*150)}{(109*0.8)+(205*0.5)}$

$T=90.2^{\circ}C$

Therefore, the temperatures at which the metals are joined is $90.2^{\circ}C$

6 0

2 years ago