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2 years ago

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roblem 10: In an adiabatic process oxygen gas in a container is compressed along a path that can be described by the following p

ressure p, in atm, as a function of volume V, in liters: p = p0 V-6/5. Here p0 is a constant of units atm⋅L6/5. show answer Incorrect Answer 50% Part (a) Write an expression for the work W done on the gas when the gas is compressed from a volume Vi to a volume Vf.

1 answer:

miskamm [114]2 years ago

3 0

Answer:

W= -2.5 (p₁*0.0012) joules

Explanation:

Given that p₀= initial pressure, p₁=final pressure, Vi= initial volume=0 and Vf=final volume= 6/5 liters where p₁=p₀ then

In adiabatic compression, work done by mixture during compression is

W= $\int\limits^f_i {p} \, dV$ where f= final volume and i =initial volume, p=pressure

p can be written as p=K/V^γ where K=p₀Vi^γ =p₁Vf^γ

W= $\int\limits^f_i {K/V^} \, dV$

W= K/1-γ ( 1/Vf^γ-1 - 1/Vi^γ-1)

W=1/1-γ (p₁Vf-p₀Vi)

W= 1/1-1.40 (p₁*6/5 -p₀*0)

W= -2.5 (p₁*6/5*0.001) changing liters to m³

W= -2.5 (p₁*0.0012) joules

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Intensity: Radiation of a single frequency reaches the upper atmosphere of the earth with an intensity of 1350 W/m2. What is the

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Answer:

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Explanation:

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Now,

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On substituting the values in the above formula, we get

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1 year ago

Suppose we were to attempt to use a similar machine to measure the charge-to-mass ratio of protons, instead. Suppose, for simpli

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Answer:

In summary, the biggest difference in the experiment is that the proton mass is much more than the electron mass, so the voltages used are high and very dangerous.

Explanation:

The machine to measure the ratio of charge / mass of the electron, has two parts: a part where it accelerates the electrons in an electric field and a second section to charter the beam and measure its radius of curvature calculated from here the q / m ratio

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The mass of the protons is much greater than the mass of the electrons, which introduced a significant difference in the excrement, since similar electric fields the speed of the protons is much less than the speed of the electrons, so the magnetic field through which the voters pass to have equivalent deflations in many cases this small values of the magnetic field are not desirable due to the interference of the Earth's magnetic field.

Another solution could be to increase the electric field to have the protons with speeds similar to the electors, this possibility is not easy either, because the field of trunking of more than 5000 V would be needed, which are very dangerous.

In summary, the biggest difference in the experiment is that the proton mass is much more than the electron mass, so the voltages used are high and very dangerous.

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2 years ago

Water evaporating from a pond does so as if it were diffusing across an air film 0.15 cm thick. The diffusion coefficient of wat

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Answer:

The water level will drop by about 1.24 cm in 1 day.

Explanation:

Here Mass flux of water vapour is given as

$j_{H_2O}=\frac{D}{l} \bigtriangleup c$

where

$j_{H_2O}$ is the mass flux of the water which is to be calculated.

D is diffusion coefficient which is given as $0.25 cm^2/s$

l is the thickness of the film which is 0.15 cm thick.
$\bigtriangleup c$ is given as

$\bigtriangleup c= \frac{P_{sat}-P_a}{RT}$

In this

$P_{sat}$ is the saturated water pressure, which is look up from the saturated water property at 20°C and 0.5 saturation given as 2.34 Pa

$P_a$ is the air pressure which is given as 0.5 times of $P_{sat}$

R is the universal gas constant as $8.314 kJ/kmol-K$

T is the temperature in Kelvin scale which is $20+273= 293K$

By substituting values in the equation

$\bigtriangleup c= \frac{P_{sat}-P_a}{RT} \\ \bigtriangleup c= \frac{P_{sat}-0.5P_{sat}}{RT} \\ \bigtriangleup c= \frac{0.5P_{sat}}{RT} \\ \bigtriangleup c= \frac{0.5 \times 2.34}{8.314 \times 293} \\\bigtriangleup c= 0.48 mol/m^3$

Converting $\bigtriangleup c$ into $cm^3/cm^3$

As 1 mole of water 18 $cm^3$ so

$\bigtriangleup c= 0.48 mol/m^3 \\ \bigtriangleup c= 0.48 \times 18 \times 10^{-6} cm^3/cm^3 \\ \bigtriangleup c= 8.64 \times 10^{-6} cm^3/cm^3$

Putting this in the equation of mass flux equation gives

$j_{H_2O}=\frac{D}{l} \bigtriangleup c \\ j_{H_2O}=\frac{0.25}{0.15} \times 8.64 \times 10^{-6} \\ j_{H_2O}=14.4 \times 10^{-6} cm/s$

For calculation of water level drop in a day, converting mass flux as

$j_{H_2O}=14.4 \times 10^{-6} \times 24 \times 3600 cm/day\\ j_{H_2O}=1.24 cm/day$

So the water level will drop by about 1.24 cm in 1 day.

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2 years ago

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Answer:195 J

Explanation:

Given

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maximum height reached by ball $h=8\ m$

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$M.E._1=(P.E.+K.E.)_1$

$M.E._1=(mgh)_1+(\frac{1}{2}mv^2)_1$

considering hand to be datum so h_1=0[/tex]

so Potential energy at ground is zero

$M.E._1=\frac{1}{2}\times m\times (15)^2$

$M.E._1=6.41\ J$

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$(M.E.)_2=(P.E.+K.E.)_2$

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$(M.E.)_2=0.0570\times 9.8\times 8$

$(M.E.)_2=4.46\ J$

Decrease in Mechanical energy

$(M.E.)_1-(M.E.)_2=6.41-4.46$

$(M.E.)_1-(M.E.)_2=1.95\ J$

3 0

2 years ago