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Tom [10]
2 years ago
8

3. A sample of argon of mass 6.56 g occupies 18.5 dm? at 305 K. (a) Calculate the work done when the gas expands isothermally ag

ainst a constant external pressure of 7.7 kPa until its volume has increased by 2.5 dm'. (b) Calculate the work that would be done if the same expansion occurred reversibly.
Physics
1 answer:
aleksandr82 [10.1K]2 years ago
4 0

Answer:

(a) W=-19.25J

(b) W=-52.8J

Explanation:

Hello.

(a) In this case, since the initial volume is 18.5 dm³ and the final volume is 21 dm³ (18.5 +2.5), we can compute the work at constant pressure as shown below:

W=-P\Delta V=-7.7kPa*\frac{1000Pa}{1kPa} (21dm^3-18.5dm^3)*\frac{1m^3}{1000dm^3}\\ \\W=-19.25J

Which is negative as it expands against the given pressure.

(b) Moreover, of the process is carried out reversibly, the pressure can change, therefore, we need to compute the work via:

W=nRTln(\frac{V_1}{V_2} )

Whereas the moles are computed from the given mass of argon:

n=6.56g*\frac{1mol}{39.95g}=0.164mol

Thus, the work is:

W=0.164mol*8.314\frac{J}{mol*K} *305Kln(\frac{18.5dm^3}{21dm^3} )\\\\W=-52.8J

Regards.

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satela [25.4K]

Answer:

Option B

Explanation:

The phase difference is found by subtracting the 2.3m for the receiver from the other speaker which is 2.9m hence

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3 0
2 years ago
The spectrum of Star A has an absorption line of hydrogen at 660.0 nm. The spectrum of Star B has an absorption line at 666 nm.
rjkz [21]

Answer:

The stars are moving away from us.

Explanation:

The observed wavelengths of hydrogen transition for stars A and B (660.0 nm and 666 nm respectively) are greater than that observed in the laboratory (656.2 nm). The observed long wavelengths for the stars means that the light from the stars is red-shifted.

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2 years ago
The water level in a tank is 20 m above the ground. a hose is connected to the bottom of the tank, and the nozzle at the end of
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Answer:

P_(pump) = 98,000 Pa

Explanation:

We are given;

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ρgh_1 + P_(pump) = ρgh_2

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P_(pump) = ρg(h_2 - h_1)

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P_(pump) = 1000•9.8(30 - 20)

P_(pump) = 98,000 Pa

5 0
2 years ago
jesse is swinging miguel in a circle at a tangential speed of 3.50 m/s. if the radius of the circle is 0.600 m and miguel has a
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Centripetal acceleration = (speed)² / (radius) .

Force = (mass) · (acceleration)

Centripetal force = (mass) · (speed)² / (radius) .

                             = (11 kg) · (3.5 m/s)² / (0.6 m)

                             = (11 kg) · (12.25 m²/s²) / (0.6 m)

                             =  (11 · 12.25) / 0.6  kg-m/s²

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That's the tension in Miguel's arm or leg or whatever part of his body
Jesse is swinging him by.  It's the centripetal force that's needed in
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6 0
2 years ago
Read 2 more answers
A weather balloon is rising vertically from a launching pad on the ground. A technician standing 300 feet from the launching pad
avanturin [10]

Answer:

\dfrac{dh}{dt} =5\ ft/s

Explanation:

Let

h = height of balloon (in feet).

θ = angle made with line of sight and ground (in radians).

h = 300  tanθ

\dfrac{dh}{d\theta } = 300 sec^2\theta

now  \dfrac{dh}{dt} can be written as

\dfrac{dh}{dt} =\dfrac{dh}{d\theta }\times \dfrac{d\theta }{dt}

\dfrac{d\theta }{dt} = \dfrac{1}{120}\at \ \theta =\dfrac{\pi}{4}

When θ = π/4,

\dfrac{dh}{d\theta } = 300 sec^2\theta

\dfrac{dh}{d\theta } = 600

\dfrac{dh}{dt} =\dfrac{dh}{d\theta }\times \dfrac{d\theta }{dt}

\dfrac{dh}{dt} =600\times \dfrac{1}{120}

\dfrac{dh}{dt} =5\ ft/s

5 0
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