♡ GuYs diG cRaZy cHiCks ♡ IvE naMeD my aLtEr Ego AJ

Determine the average value of the translational kinetic energy of the molecules of an ideal gas at (a) 27.8°C and (b) 143°C. Wh

Alinara [238K]

Answer:

a) $k_{avg}=6.22\times 10^{-21}$

b) $k_{avg}=8.61\times 10^{-21}$

c) $k_{mol}=3.74\times 10^{3}J/mol$

d) $k_{mol}=5.1\times 10^{3}J/mol$

Explanation:

Average translation kinetic energy ( $k_{avg}$ ) is given as

$k_{avg}=\frac{3}{2}\times kT$ ....................(1)

where,

k = Boltzmann's constant ; 1.38 × 10⁻²³ J/K

T = Temperature in kelvin

a) at T = 27.8° C

or

T = 27.8 + 273 = 300.8 K

substituting the value of temperature in the equation (1)

we have

$k_{avg}=\frac{3}{2}\times 1.38\times 10^{-23}\times 300.8$

$k_{avg}=6.22\times 10^{-21}J$

b) at T = 143° C

or

T = 143 + 273 = 416 K

substituting the value of temperature in the equation (1)

we have

$k_{avg}=\frac{3}{2}\times 1.38\times 10^{-23}\times 416$

$k_{avg}=8.61\times 10^{-21}J$

c ) The translational kinetic energy per mole of an ideal gas is given as:

$k_{mol}=A_{v}\times k_{avg}$

here $A_{v}$ = Avagadro's number; ( 6.02×10²³ )

now at T = 27.8° C

$k_{mol}=6.02\times 10^{23}\times 6.22\times 10^{-21}$

$k_{mol}=3.74\times 10^{3}J/mol$

d) now at T = 143° C

$k_{mol}=6.02\times 10^{23}\times 8.61\times 10^{-21}$

$k_{mol}=5.1\times 10^{3}J/mol$

8 0

3 years ago

6. An earthquake releases two types of traveling seismic waves, called transverse and longitudinal waves. The average speed of t

zubka84 [21]

Answer:

The distance away the center of the earthquake is 1083.24 km.

Explanation:

Given that,

Speed of transverse wave = 9.1\ km/s

Speed of longitudinal wave = 5.7 km/s

Time = 71 sec

We need to calculate the distance of transverse wave

Using formula of distance

$d=v\times t$

$d=9.1\times t$ ....(I)

The distance of longitudinal wave

$d=5.7\times (t+71)$ ....(II)

From the first equation

$t=\dfrac{d}{9.1}$

Put the value of t in equation (II)

$d =5.7\times(\dfrac{d}{9.1}+71)$

$\dfrac{9.1d-5.7d}{9.1}=71\times5.7$

$d0.3736=404.7$

$d =1083.24\ km$

Hence, The distance away the center of the earthquake is 1083.24 km.

6 0

2 years ago

a tms (transcranial magnetic stimulation) device creates very rapidly changing magnetic fields. the field near a typical pulsed-

Digiron [165]

Part A : 4.33 Volt.

Part B : 4764 Amperes.

Given Data :

Field near a typical pulsed-field machine rises from 0 T to 2.5 T in 200 μs.

Axis of his 2.1-cm-diameter wedding band is parallel to the field.

Part A :

induced emf = rate of change of magnetic flux

= d (B A )/dt

= A dB/dt

= { π (2.1 x 10^-2 / 2)^2) (2.5) } / (200 x 10^-6)

E = 4.33 Volt

Part B :

R = rho L / A

R = { (6.2 x 10^-8) ( π x 0.021) } / (4.5 x 10^-6)

R = 9.10 x 10^-4 Ohm

We know, (I = V/ R)

I = 4.33 Volts / 9.10 x 10^-4 Ohm

I = 4764 Amperes

What exactly is the function of TMS?

Transcranial magnetic stimulation (TMS) is a noninvasive procedure that uses magnetic fields to stimulate nerve cells in the brain to alleviate depression symptoms. TMS is usually used when other depression treatments have failed.

Find more on TMS at : brainly.com/question/13851741

#SPJ4

➤ Although your question isn't complete, I may have assumed that you were referring to this specific question.

The complete question is :

A TMS (transcranial magnetic stimulation) device creates very rapidly changing magnetic fields. The field near a typical pulsed-field machine rises from 0 T to 2.5 T in 200 μs. Suppose a technician holds his hand near the device so that the axis of his 2.1-cm-diameter wedding band is parallel to the field.

Part A: What emf is induced in the ring as the field changes? Express your answer to two significant figure E= ___ unit_

Part B: If the band is made of a gold alloy with resistivity 6.2×10−8Ω⋅m and has a cross-section area 4.5 mm2 , what is the induced current? Express your answer to two significant figures and include the appropriate units. I= ___ units__

6 0

1 year ago

Estimate how long a 2500 W electric kettle would take to boil away 1.5 Kg of water . The specific latent heat of vaporization of

andrew-mc [135]

The time it would take a 2500 W electric kettle to boil away 1.5 Kg of water is 2400 seconds

<h3>How to calculate the time</h3>

Use the formula:

Power × time = mass × specific heat

Given mass = 1. 5kg

Specific latent heat of vaporization = 4000000 J/ Kg

Power = 2500 W

Substitute the values into the formula

Power × time = mass × specific heat

2500 × time = 1. 5 × 4000000

Make 'time' the subject

time = 1. 5 × 4000000 ÷ 2500 = 6000000 ÷ 2500 = 2400 seconds

Therefore, the time it would take a 2500 W electric kettle to boil away 1.5 Kg of water is 2400 seconds.

Learn more about specific latent heat of vaporization:

https://brainly.in/question/1580957

#SPJ1

8 0

1 year ago

The energy transferred from the water in the kettle to the surroundings in 2 hours is

nignag [31]

The Last one ☝️ I think

6 0

2 years ago