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ss7ja [257]
2 years ago
6

In this problem you are to consider an adiabaticexpansion of an ideal diatomic gas, which means that the gas expands with no add

ition or subtraction of heat.
Assume that the gas is initially at pressure p0, volume V0, and temperature T0. In addition, assume that the temperature of the gas is such that you can neglect vibrational degrees of freedom. Thus, the ratio of heat capacities is γ=Cp/CV=7/5.

Note that, unless explicitly stated, the variable γ should not appear in your answers--if needed use the fact that γ=7/5 for an ideal diatomic gas.

Part A

Find an analytic expression for p(V), the pressure as a function of volume, during the adiabatic expansion.

Express the pressure in terms of V and any or all of the given initial values p0, T0, and V0.

Correct

p(V) =
p0(V0V)75

Part B

At the end of the adiabatic expansion, the gas fills a new volume V1, where V1>V0. Find W, the work done by the gas on the container during the expansion.

Express the work in terms of p0, V0, and V1. Your answer should not depend on temperature.

Part C

Find ΔU, the change of internal energy of the gas during the adiabatic expansion from volume V0 to volume V1.

Express the change of internal energy in terms of p0, V0, and/or V1.

Need Part B and C

Physics
1 answer:
LenKa [72]2 years ago
6 0

Answer:

a)

The analytic expression for p(V) is P(V) = P_{0}(\frac{V_{0}}{V})^{7/5}

b)

The expression of the work in terms of p0 ,V0,V1 is

          W= \frac{5}{2} P_{0}[V_{0}-V_{0}^{7/5}V_{1}^{-2/5}]

c)

The change in internal energy of the gas

     \delta U = \frac{5}{2}P_{0}[V_{0}^{7/5}V_{1}^{-2/5}-V_{0}]

Explanation:

The explanation is shown on the first, second and third

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Answer:

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Init. A
Gekata [30.6K]

Answer:

v = 1/3 m / s = 0.333 m / s

in the direction of the truck

Explanation:

The average speed is defined by the variation of the position between the time spent

           v = Δx / Δt

since the position is a vector we must add using vectors, we will assume that the displacement to the right is positive, the total displacement is

           Δx = 20 - 15 +20

           Δx = 25 m

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A permeability test was run on a compacted sample of dirty sandy gravel. The sample was 175 mm long and the diameter of the mold
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Answer:

(a). The coefficient of permeability is 8.6\times10^{-3}\ cm/s.

(b). The seepage velocity is 0.0330 cm/s.

(c). The discharge velocity during the test is 0.0187 cm/s.

Explanation:

Given that,

Length = 175 mm

Diameter = 175 mm

Time = 90 sec

Volume= 405 cm³

We need to calculate the discharge

Using formula of discharge

Q=\dfrac{V}{t}

Put the value into the formula

Q=\dfrac{405}{90}

Q=4.5\ cm^3/s

(a). We need to calculate the coefficient of permeability

Using formula of coefficient of permeability

Q=kiA

k=\dfrac{Q}{iA}

k=\dfrac{Ql}{Ah}

Where, Q=discharge

l = length

A = cross section area

h=constant head causing flow

Put the value into the formula

k=\dfrac{4.5\times175\times10^{-1}}{\dfrac{\pi(175\times10^{-1})^2}{4}\times38}

k=8.6\times10^{-3}\ cm/s

The coefficient of permeability is 8.6\times10^{-3}\ cm/s.

(c). We need to calculate the discharge velocity during the test

Using formula of discharge velocity

v=ki

v=\dfrac{kh}{l}

Put the value into the formula

v=\dfrac{8.6\times10^{-3}\times38}{17.5}

v=0.0187\ cm/s

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(b). We need to calculate the volume of solid in the ample

Using formula of volume

V_{s}=\dfrac{M_{s}}{V_{s}}

Put the value into the formula

V_{s}=\dfrac{4950\times10^{-3}}{2710}

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We need to calculate the volume of the soil specimen

Using formula of volume

V=A\times L

Put the value into the formula

V=\dfrac{\pi(17.5)^2}{4}\times17.5

V=4209.24\ cm^3

We need to calculate the volume of the voids

V_{v}=V-V_{s}

Put the value into the formula

V_{v}=4209.24-1826.56

V_{v}=2382.68\ cm^3

We need to calculate the seepage velocity

Using formula of velocity

Av=A_{v}v_{s}

v_{s}=\dfrac{Av}{A_{v}}

v_{s}=\dfrac{V}{V_{v}}\times v

Put the value into the formula

v_{s}=\dfrac{4209.24}{2382.68}\times0.0187

v_{s}=0.0330\ cm/s

The seepage velocity is 0.0330 cm/s.

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(b). The seepage velocity is 0.0330 cm/s.

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8 0
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Alenkasestr [34]

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Divide each side by  4.9 :   0.92 / 4.9 = T²

Take the square root
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