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Elenna [48]
2 years ago
10

Suppose that we use a heater to boil liquid nitrogen (N2 molecules). 4480 J of heat turns 20 g of liquid nitrogen into gas. Note

that the latent heat is equal to the change in enthalpy, and that liquid nitrogen boils at 77 K. The system is kept at a constant pressure of 1 atm. 20) Assuming that you can treat the gas as ideal gas and that the volume of the liquid compute the binding energy of a nitrogen molecule in the liquid. (the binding energy is the difference in internal energy per molecule between the liquid and gas) approximately zero,
a. 9.4 x 10-21 J
b. 3.8 х 1027 J
c. 4.2 x 10-18 J
d. 10-20 J e. 2.1 x 10-19 J
Physics
1 answer:
love history [14]2 years ago
4 0

Answer:

The energy is 9.4\times10^{-21}\ J

(a) is correct option

Explanation:

Given that,

Energy = 4480 j

Weight of nitrogen = 20 g

Boil temperature = 77 K

Pressure = 1 atm

We need to calculate the internal energy

Using first law of thermodynamics

Q=\Delta U+W

Q=\Delta U+nRT

Put the value into the formula

4480=\Delta U+\dfrac{20}{28}\times8.314\times77

\Delta U=4480-\dfrac{20}{28}\times8.314\times77

\Delta U=4022.73\ J

We need to calculate the number of molecules in 20 g N₂

Using formula of number of molecules

N=n\times \text{Avogadro number}

Put the value into the formula

N=\dfrac{20}{28}\times6.02\times10^{23}

N=4.3\times10^{23}

We need to calculate the energy

Using formula of energy

E=\dfrac{\Delta U}{N}

Put the value into the formula

E=\dfrac{4022.73}{4.3\times10^{23}}

E=9.4\times10^{-21}\ J

Hence, The energy is 9.4\times10^{-21}\ J

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3 0
3 years ago
A rod has a radius of 10 mm is subjected to an axial load of 15 N such that the axial strain in the rod is ????௫ = 2.75*10-6, de
EleoNora [17]

Answer:

Knowing we only have one load applied in just one direction we have to use the Hooke's law for one dimension

ex = бx/E

бx = Fx/A = Fx/πr^{2}

Using both equation and solving for the modulus of elasticity E

E = бx/ex = Fx / πr^{2}ex

E = \frac{15}{pi (10 * 10^{-3})^{2} * 2.75 * 10^{-6}    } = 17.368 * 10^{9} Pa = 17.4 GPa

Apply the Hooke's law for either y or z direction (circle will change in every direction) we can find the change in radius

ey = \frac{1}{E} (бy - v (бx + бz)) = -\frac{v}{E}бx

= \frac{vFx}{Epir^{2} } = \frac{0.23 * 15}{pi (10 * 10^{-3)^{2} } * 17.362 * 10^{9}  } = -0.63 *10^{-6}

Finally

ey = Δr / r

Δr = ey * r = 10 * -0.63* 10^{-6} mm = -6.3 * 10^{-6} mm

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3 years ago
A solid 0.4550 kg ball rolls without slipping down a track toward a vertical loop of radius R = 0.6750 m . What minimum translat
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Answer:

u = 3.35 m/s

Explanation:

given,

mass , m = 0.455 kg  

R = 0.675 m

Height of Loop = 1.021 m

the speed required at the top of loop be v

equating the force vertically

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9.81 =\dfrac{v^2}{0.675}

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using conservation of energy

\dfrac{1}{2}mu^2 + \dfrac{1}{2}I\omega^2 + m g h = \dfrac{1}{2}I\omega^2 + \dfrac{1}{2}mv^2+ m g (2 R)

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\dfrac{1}{2}mu^2 + \dfrac{1}{2}(\dfrac{2}{5}mr^2)(\dfrac{u}{r})^2 + m g h = \dfrac{1}{2}(\dfrac{2}{5}mr^2)(\dfrac{v}{r})^2 + \dfrac{1}{2}mv^2+ m g (2 R)

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u² = 11.2156

u = 3.35 m/s

the initial  speed is 3.35 m/s

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