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bekas [8.4K]
2 years ago
10

Compressed gases aren't ideal. Let's consider a gas that's non-ideal only because the volume available to each of the N molecule

s is reduced because each other molecule occupies volume v. Instead of pV=NkT, we get: p(V-Nb)=NkT. Let b=7 × 10^-29 m3. Let's look at 3 moles of this gas at T=300K starting in 0.001 m3 volume.
1) What's the initial value of the pressure? p_initial = ?

2) The gas expands isothermally to 0.002 m3. What's the final pressure? p_final = ?

3) How much work did the gas do in this isothermal expansion? W = ?
Physics
1 answer:
Colt1911 [192]2 years ago
7 0

Answer:

8563732.58906 Pa

3992793.23326 Pa

5708.00923 J

Explanation:

V = Volume

N = Number of molecules = 3\times 6.023\times 10^{23}

T = Temperature = 300 K

b = 7\times 10^{-29}\ m^3

k_ = Boltzmann constant = 1.38\times 10^{-23}\ J/K

P = Pressure

We have the equation

P(V-Nb)=NkT\\\Rightarrow P=\dfrac{NkT}{V-Nb}\\\Rightarrow P=\dfrac{3\times 6.023\times 10^{23}\times 1.38\times 10^{-23}\times 300}{0.001-3\times 6.023\times 10^{23}\times 7\times 10^{-29}}\\\Rightarrow P=8563732.58906\ Pa

The pressure is 8563732.58906 Pa

For isothermal expansion

P_1(V_1-Nb)=P_2(V_2-Nb)\\\Rightarrow P_2=\dfrac{P_1(V_1-Nb)}{V_2-Nb}\\\Rightarrow P_2=\dfrac{8563732.58906(0.001-3\times 6.023\times 10^{23}\times 7\times 10^{-29})}{0.002-3\times 6.023\times 10^{23}\times 7\times 10^{-29}}\\\Rightarrow P_2=3992793.23326\ Pa

The pressure is 3992793.23326 Pa

Work done is given by

dw=Pdv\\\Rightarrow W=\int_{v_1}^{v_2}\dfrac{NkT}{V-Nb}dv\\\Rightarrow W=NkTln\dfrac{V_2-Nb}{V_1-Nb}\\\Rightarrow W=3\times 6.023\times 10^{23}\times 1.38\times 10^{-23}\times 300ln\dfrac{0.002-3\times 6.023\times 10^{23}\times 7\times 10^{-29}}{0.001-3\times 6.023\times 10^{23}\times 7\times 10^{-29}}\\\Rightarrow W=5708.00923\ J

The work done is 5708.00923 J

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Answer:

The answer to the question is

3340800 m far

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Time of acceleration = 8 minutes

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S = distance traveled, m

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A 100-kg tackler moving at a speed of 2.6 m/s meets head-on (and holds on to) an 92-kg halfback moving at a speed of 5.0 m/s. Pa
DIA [1.3K]

Given that,

Mass of trackler, m₁ = 100 kg

Speed of trackler, u₁ = 2.6 m/s

Mass of halfback, m₂ = 92 kg

Speed of halfback, u₂ = -5 m/s (direction is opposite)

To find,

Mutual speed immediately after the collision.

Solution,

The momentum of the system remains conserved in this case. Let v is the mutual speed after the collision. Using conservation of momentum as :

m_1u_1+m_2u_2=(m_1+m_2)V\\\\V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}\\\\V=\dfrac{100\times 2.6+92\times (-5)}{(100+92)}\\\\V=-1.04\ m/s

So, the mutual speed immediately after the collision is 1.04 m/s but in opposite direction.

3 0
3 years ago
A hydrogen discharge lamp emits light with two prominent wavelengths: 656 nm (red) and 486 nm (blue). The light enters a flint-g
mezya [45]

Answer:

The angle between the red and blue light is 1.7°.

Explanation:

Given that,

Wavelength of red = 656 nm

Wavelength of blue = 486 nm

Angle = 37°

Suppose we need to find the angle between the red and blue light as it leaves the prism

n_{r}=1.572

n_{b}=1.587

We need to calculate the angle for red wavelength

Using Snell's law,

n_{r}\sin\theta_{i}=n_{a}\sin\theta_{r}

Put the value into the formula

1.572\sin37=1\times\sin\theta_{r}

\theta_{r}=\sin^{-1}(\dfrac{1.572\sin37}{1})

\theta_{r}=71.0^{\circ}

We need to calculate the angle for blue wavelength

Using Snell's law,

n_{b}\sin\theta_{i}=n_{a}\sin\theta_{b}

Put the value into the formula

1.587\sin37=1\times\sin\theta_{b}

\theta_{b}=\sin^{-1}(\dfrac{1.587\sin37}{1})

\theta_{b}=72.7^{\circ}

We need to calculate the angle between the red and blue light

Using formula of angle

\Delta \theta=\theta_{b}-\theta_{r}

Put the value into the formula

\Delta \theta=72.7-71.0

\Delta \theta=1.7^{\circ}

Hence, The angle between the red and blue light is 1.7°.

8 0
3 years ago
2. An 873 kg dragster accelerates at a rate of 44.6 m/s during a race.
Ira Lisetskai [31]

Answer:

<h3>38,673.9N</h3>

Explanation:

According to newton's second law:

Force = mass * acceleration

Given

Mass = 873kg

acceleration = 44.66m/s²

Magnitude of the force is expressed as;

F = ma

F = 873 * 44.6

F = 38,673.9N

<em>Hence the magnitude of the net force exerted on the dragster during this time is 38,673.9N</em>

8 0
2 years ago
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