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hodyreva [135]
3 years ago
15

What is the temperature change of 1.0 mol of a monotomic gas if its thermal energy is increased by 1.0 J ? Express your answer i

n kelvins.
Physics
1 answer:
natali 33 [55]3 years ago
7 0

Answer: 0.08K

Explanation:

When temperature changes, the corresponding change in thermal energy of a gas is given by:

ΔE (thermal) = 3/2nRΔT

Defining the parameters:

ΔE (thermal) = Increase in thermal energy of the mono atomic gas = 1.0J

n = number of moles of the gas = 1.0mol

R = Ideal gas constant = 8.314J/mol/K

ΔT = change in temperature. This is what we need to find.

Rearranging the equation to make ΔT subject of the formula,

ΔT= 2 x ΔE (thermal) / (3 x n x R)

Therefore, ΔT = 2 x 1.0J / (3 x 1.0mol x 8.314J/mol/K)

ΔT = 2.0J / 24.942J/K

ΔT = 0.0802K

ΔT = 0.08K

The temperature change of 1.0mol of a monoatomic gas if its thermal energy is increased by 1.0J is 0.08K.

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Answer:

<u>0.04 °C⁻¹</u>

Explanation:

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<u />

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⇒ α = 5 / 5 × 50

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⇒ Area Expansivity = 2 × Linear Expansivity

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Answer:

-2.478

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Explanation:

Angular frequency of spring in harmonic motion is given by?

ω = √(k/m)

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ω = √4.54

ω = 2.13 s^-1

If at t=0 the mass is in negative amplitude (x = -A = -2.48 m) then we describe the position with negative cosine

x(t) = -A * cos(ωt)

x(t) = -2.48 * cos(2.13 * 1)

x(t) = -2.48 * 0.9993

x(t) = -2.478

Velocity and acceleration are 1st and 2nd derivative of position

b)

v(t) = Aω * sin(ωt)

v(t) = 2.48 * 2.13 * sin(2.13 * 1)

v(t) = 5.282 * sin2.13

v(t) = 5.282 * 0.03717

v(t) = 0.379 m/s

c)

a(t) = Aω^2 * cos(ωt)

a(t) = 2.48 * 2.12² * cos(2.13 * 1)

a(t) = 2.48 * 4.494 * cos2.13

a(t) = 11.15 * 0.9993

a(t) = 11.14 m/s²

d)

F = -k * x(t)

F = -10 * -2.478

F = 24.78 N

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Arte-miy333 [17]

the sodium chloride will be a crystal

it will have a giant crystal lattice

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3 years ago
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