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3 years ago

What is absolute zero? What is the temperature of absolute zero on the Kelvin and Celsius scales?

Physics

2 answers:

vredina [299]3 years ago

7 0

Answer:

0 Kelvin

Explanation:

(Educere/Founder's Education Answer)

salantis [7]3 years ago

6 0

Answer:

Absolute zero = 0 K or - 273°C

Explanation:

Absolute zero :

When the entropy and enthalpy of the ideal system reach at the minimum value then the temperature at that condition is known as absolute zero condition.

Absolute temperature is the minimum temperature in the temperature scale.The value of absolute zero is 0 K.

We know that

$\dfrac{C-0}{100}=\dfrac{K-273}{100}=\dfrac{F-32}{180}$

F=Temperature in Fahrenheit scale

K=Temperature in Kelvin scale

C=Temperature in degree Celsius scale

When K = 0

$\dfrac{C-0}{100}=\dfrac{K-273}{100}$

$\dfrac{C-0}{100}=\dfrac{0-273}{100}$

C= - 273°C

Absolute zero = 0 K or - 273°C

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Because noble gases has full of 8 electron in her external cell......

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3 years ago

Read 2 more answers

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

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

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

A 7750 kg space probe, moving nose-first toward Jupiter at 179 m/s relative to the Sun, fires its rocket engine, ejecting 72.0 k

Reika [66]

Answer:

179.47m/s

Explanation:

Using the law of conservation of momentum

m1u1 + m2u2 = (m1+m2)v

m1 and m2 are the masses

u1 and u2 are the initial velocities

v is the final velocity

Substitute

7750(179)+72(230) = (7750+72)v

1,387,250+16560 = 7822v

1,403,810 = 7822v

v = 1,403,810/7822

v= 179.47m/s

Hence the final velocity of the probe is 179.47m/s

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3 years ago

Give two examples of events that show that the speed of sound is very much slower than the speed of light

jeka94

Answer:

1) Lightning, you see the lightning first and then hear the thunder.

2)When a person far away from you hits a ball with a bat, you can see them striking the ball first and then you will hear the sound of ball striking against the bat.

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Hunter-Best [27]

Answer:

Explanation: see attachment below

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3 years ago