The magnetic field of a magnet is strongest

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

2 years ago

Kayla and her friends are setting up chairs for a school play each row will contain the same number of chairs Kayla knows that t

LUCKY_DIMON [66]

Answer:

96=8*c

Explanation:

4 0

2 years ago

What is the photoelectric effect?

n200080 [17]

the emission of electrons or other free carriers when light shines on a material.

4 0

3 years ago

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Two protons in an atomic nucleus are typically separated by a distance of 2 ✕ 10-15 m. The electric repulsion force between the

castortr0y [4]

Answer:

The magnitude of the electric force between the to protons will be 57.536 N.

Explanation:

We can use Coulomb's law to find out the force, in scalar form, will be:

$F \ = \ \frac{1}{4 \pi \epsilon_0 } \frac{q_1 q_2}{d^2}$ .

Now, making the substitutions

$d \ = \ 2.00 * 10 ^{-15} \ m$ ,

$q_1 = q_2 = 1.60 * 10 ^ {-19} \ C$ ,

$\frac{1}{4\pi\epsilon_0}=8.99 * 10^9 \frac{Nm^2}{C^2}$ ,

we can find:

$F \ = \ 8.99 * 10^9 \frac{Nm^2}{C^2} \frac{(1.60 * 10 ^ {-19} \ C)^2}{(2.00 * 10 ^{-15} \ m)^2}$ .

$F \ = 57.536 N$ .

Not so big for everyday life, but enormous for subatomic particles.

4 0

2 years ago

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Consider a skateboarder who starts from rest at the top of ramp that is inclined at an angle of 18.0 ∘ to the horizontal.

kotegsom [21]

Answer:

Explanation:

v= u + at

v is final velocity , u is initial velocity . a is acceleration and t is time

Initial velocity u = 0 . Putting the given values in the equation

v = 0 + g sin 18 x 3.5

= 10.6 m /s

3 0

2 years ago