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BaLLatris [955]
3 years ago
10

Kinetic energy kx is 1/2mvx2. evaluate the mean square kinetic energy < kx2 >in one dimension for n2 molecules at 25°c (hi

nt: in the derivation of < kx2 > use f (vx.) )
Engineering
1 answer:
mash [69]3 years ago
3 0

The <em>mean kinetic</em> energy of the N₂ molecules at 25 °C is 3593.519 joules per mole.

<h3>How to estimate the mean square kinetic energy associated to gas molecules</h3>

Let suppose that the <em>gas</em> sample behaves ideally. The Graham's law establishes a connection between the <em>ideal gas</em> model and the <em>kinetic</em> theory of gases.

In this question we need to use this law to estimate the <em>average</em> <em>kinetic</em> energy (<em>Kₐ</em>), in joules per mole, of a <em>diatomic</em> gas (<em>N₂</em>), which is defined by the following expression:

K_{a} = \frac{3}{2}\cdot R_{u}\cdot T   (1)

Where:

  • R_{u} - Ideal gas constant, in joules per mole-Kelvin.
  • T - Temperature, in Kelvin

If we know that R_{u} = 8.314\,\frac{J}{mol\cdot K} and T = 288.15\,K, then the average kinetic energy is:

K_{a} = \frac{3}{2}\cdot \left(8.314\,\frac{J}{mol\cdot K} \right) \cdot (288.15\,K)

K_{a} = 3593.519\,\frac{J}{mol}

The <em>mean kinetic</em> energy of the N₂ molecules at 25 °C is 3593.519 joules per mole. \blacksquare

To learn more on kinetic theory of gases, we kindly invite to check this verified question: brainly.com/question/15064212

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Pani-rosa [81]

Answer:

4) a) s = +4\,mi, b) \Delta s = 2\,mi, 5) a) s = -2\,mi, b) \Delta s = 10\,mi

Explanation:

4) Let assume that distance with respect to origin is positive at east, while distance is negative at west.

a) The initial displacement is:

s = +4\,mi

b) The distance travelled is:

\Delta s = 2\,mi

\Delta s = 2\,mi

5) a) The final displacement is:

s = -2\,mi

b) The total distance travelled is:

\Delta s = 2\,mi + 6\,mi + 2\,mi

\Delta s = 10\,mi

5 0
4 years ago
If the resistance in a series circuit is 12 K Ohms, and the amperage is 0. 0063 amps, what is the voltage?
kkurt [141]

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The resistance “seen” by the 12-Volt battery is 500 ohms . The total resistance is the sum of the resistors R1 and R2. Each resistor will drop a calculable amount of voltage. This voltage is given by Ohm's Law.

3 0
2 years ago
What is
Yuri [45]

Answer:

1

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6 0
2 years ago
Read 2 more answers
1. A gas pressure difference is applied to the legs of a U-tube manometer filled with a liquid with S = 1.5. The manometer readi
julia-pushkina [17]

Answer:

1) The pressure difference is 4.207 kilopascals.

2) 2.5 pounds per square inch equals 5.093 inches of mercury and 5.768 feet of water.

Explanation:

1) We can calculate the gas pressure difference from the U-tube manometer by using the following hydrostatic formula:

\Delta P = \frac{S\cdot \rho_{w}\cdot g \cdot \Delta h}{1000} (Eq. 1)

Where:

S - Relative density, dimensionless.

\rho_{w} - Density of water, measured in kilograms per cubic meter.

g - Gravitational acceleration, measured in meters per square second.

\Delta h - Height difference in the U-tube manometer, measured in meters.

\Delta P - Gas pressure difference, measured in kilopascals.

If we know that S = 1.5, \rho_{w} = 1000\,\frac{kg}{m^{3}}, g = 9.807\,\frac{m}{s^{2}} and \Delta h = 0.286\,m, then the pressure difference is:

\Delta P = \frac{1.5\cdot \left(1000\,\frac{kg}{m^{3}} \right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.286\,m)}{1000}

\Delta P = 4.207\,kPa

The pressure difference is 4.207 kilopascals.

2) From Physics we remember that a pound per square unit equals 2.036 inches of mercury and 2.307 feet of water and we must multiply the given pressure by corresponding conversion unit: (p = 2.5\,psi)

p = 2.5\,psi\times 2.037\,\frac{in\,Hg}{psi}

p = 5.093\,in\,Hg

p = 2.5\,psi\times 2.307\,\frac{ft\,H_{2}O}{psi}

p = 5.768\,ft\,H_{2}O

2.5 pounds per square inch equals 5.093 inches of mercury and 5.768 feet of water.

4 0
4 years ago
A 4 cm diameter sphere of copper is initially at a temperature of 95 °C. It is placed in a very large water bath at time t- 0. T
avanturin [10]

Answer:112.376 s

Explanation:

Given

T_i=95^{\circ}C

T_f=35^{\circ}C

T_\infty \left ( ambient\right )=25^{\circ}C

h=400 watts/\left ( m^{2}^{\circ}C\right )

c=0.385 J/\left ( m^2^{\circ}C\right )

\rho =9 gm/cm^{3}

Using Newton's law of cooling

\frac{T_i-T_{\infty}}{T-T_{\infty}}=e^{\frac{ht}{\rho L_{c}c}}

\frac{95-25}{35-25}=e^{\frac{400\times 3\times 10^{-4}\times t}{9\times 2\times 0.385}}

7=e^{1.7316\times 10^{-2}\times t}

Taking log both side

t=112.376sec

4 0
3 years ago
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