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3 years ago

Boltzmann’s constant is 1.38066×10^−23 J/K,

Physics

1 answer:

dmitriy555 [2]3 years ago

8 0

Answer:

1.18266×10⁻²⁰ J

Explanation:

Applying,

E = (3/2)kT.................... Equation 1

Where E = kinetic energy of the gas molecule, k = Boltzmann's constant, T = Temperature

But,

PV = nRT.............. Equation 2

Where P = pressure, V = Volume, n = number of moles, R = Universal gas constant.

make T the subject of the equation

T = PV/nR............. Equation 3

Substitute equation 3 into equation 1

E = (3/2)k(PV/nR)

E = 3kPV/2nR............... Equation 4

Given: k = 1.38066×10⁻²³ J/K, V = 6.6 L = 0.0066 m³, P = 7.1 atm = (101325×7.1) N/m² = 719407.5 N/m², n = 2.9 mol, R = 8.31451 J/K.mol

Substitute into equation 4

E = (3×1.38066×10⁻²³×0.0066×719407.5)/(2×8.31451)

E = 1182.66×10⁻²³ J

E = 1.18266×10⁻²⁰ J

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What quantity of heat must be removed from 20g<br>of water at 0°C to change it to ice at 0°C?

seraphim [82]

The quantity of heat must be removed is 1600 cal or 1,6 kcal.

<h3>Explanation : </h3>

From the question we will know if the condition of ice is at the latent point. So, the heat level not affect the temperature, but it can change the object existence. So, for the formula we can use.

$\boxed {\bold {Q = m \times L}}$

If :

Q = heat of latent (cal or J )
m = mass of the thing (g or kg)
L = latent coefficient (cal/g or J/kg)

<h3>Steps : </h3>

If :

m = mass of water = 20 g => its easier if we use kal/g°C
L = latent coefficient = 80 cal/g

Q = ... ?

Answer :

$Q = m \times L \\ Q = 20 \times 80 = 1600 \: cal$

So, the quantity of heat must be removed is 1600 cal or 1,6 kcal.

<u>Subject : Physics </u>

<u>Subject : Physics Keyword : Heat of latent</u>

4 0

3 years ago

Calculate the de Broglie wavelength of an electron accelerated from rest through a potential difference of (a) 100 V, (b) 1.0 kV

forsale [732]

Answer:

(a) $\lambda=1.227\ A$

(b) $\lambda=0.388\ A$

Explanation:

Given that,

(a) An electron accelerated from rest through a potential difference of 100 V. The De Broglie wavelength in terms of potential difference is given by :

$\lambda=\dfrac{h}{\sqrt{2meV} }$