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2 years ago

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A heat engine using a diatomic ideal gas goes through the following closed cycle: ∙ Isothermal compression until the volume is h

alved. ∙ Isobaric expansion until the volume is restored to its initial value. ∙ Isochoric cooling until the pressure is restored to its initial value.

1 answer:

Alona [7]2 years ago

6 0

Answer:

a)η = 0.088

b) η = 0.5

Explanation:

a) The attached figure shows the P-V diagram for the process described in the exercise. According to that figure, the work during process 1-2 is equal to:

W(1-2) = -n*R*T1*ln(vi/vf) = -n*R*T1*ln(V/(V/2)) = -n*R*T1*ln(2)

the work during process 2-3 is equal to:

W(2-3) = nR*(T2-T1)

The work done during the 3-1 process equals zero, because the volume is constant. The specific heat for the molar specific heat equals:

cp = 7*R/2, where R is gas constant.

Qin = n*cp*(T2-T1) = 7*n*R/2*(T2-T1)

the efficiency of the cycle is equal to:

η = (W(1-2) + W(2,3))/Qin = (-n*R*T1*ln2 + n*R*(T2-T1))/(7*n*R/2*(T2-T1) = (2/7)*(1-(ln2/((T2/T1)-1)))

if we write the expression between volume and temperature, we have:

T2/T1 = v1/v2

T2/T1 = v/(v/2))

T2/T1 = 2

η = (2/7)*(1-(ln2/(2-1))) = 0.088

b)

The equation for efficiency of Carnot will be equal to:

η = 1-(T1/T2) = 1 - (1/2) = 0.5

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Answer:

0.83 m or 5.57 m

Explanation:

Destructive interference will occur when the distances from the speakers differ by 1/2 wavelength.

The length of 1 cycle of 72.4 Hz is ...

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So, the distance of the listener from speaker B is ...

3.2 m ± (4.738 m)/2 = {0.83 m, 5.57 m} . . . either of these distances

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The location could be at additional multiples of 4.738 m, but we think not. The sound intensity drops off with the square of the distance from the speaker, so identical sound waves from the speakers will sound quite different at different distances from the speakers. For best interference, the distances need to be as close to the same as possible. That will be at 3.2 m and 5.57 m.

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<em>Comment on the speed of sound</em>

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Technician a says that using a pressure transducer and lab scope is a similar process to using a vacuum gauge. technician b says

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Answer: Both Technician A and B

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There is a similar process in using a pressure transducer and lab scope to using a vacuum gauge.

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you want to compare brands of paper towels to see which holds the most liquid. the independent variable in your experiment would

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Answer: the brand of paper towel

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The pupil of the human eye can vary in diameter from 2.00 mm in bright light to 8.00 mm in dim light. The eye has a focal length

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The indicated data are of clear understanding for the development of Airy's theory. In optics this phenomenon is described as an optical phenomenon in which The Light, due to its undulatory nature, tends to diffract when it passes through a circular opening.

The formula used for the radius of the Airy disk is given by,

$y_r=1.22\frac{\lambda f}{d}$

Where,

$y_r =$ Range of the radius

$\lambda =$ wavelength

f= focal length

Our values are given by,

State 1:

$d=2.00mm = 2*10^{-3}m$

$f= 25mm = 25*10^{-3}m$

$\lambda = 750nm = 750*10^{-9}m$

State 2:

$d=8.00mm = 8*10^{-3}m$

$f= 25mm = 25*10^{-3}m$

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Replacing in the first equation we have:

$y_{r1} = 1.22\frac{(750*10^{-9})(25*10^{-3})}{2*10^{-3}}$

$y_{r1}= 11.4\mu m$

And also for,

$y_{r2} =1.22\frac{(390*10^{-9})(25*10^{-3})}{8*10^{-3}}$

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2 years ago

A 0.20-kg object attached to the end of a string swings in a vertical circle (radius = 80 cm). at the top of the circle the spee

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Answer:

Tension in the string at this position: 3.1 N.

Explanation:

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$r = 80\;\text{cm} = 0.80\;\text{m}$ .

What's the net force on the object?

The object is in a circular motion. As a result,

$\displaystyle \Sigma F = \frac{m\cdot v^{2}}{r}$ ,

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$\Sigma F$ is the net force on the object,
$m$ is the mass of the object,
$v$ is the velocity of the object, and
$r$ is the radius of the circular motion.

For this object,

$\displaystyle \Sigma F = \frac{0.20\times {4.5}^{2}}{0.80} = 5.0625\;\text{N}$ .

The output unit of net force should be standard if the unit for mass, velocity, and radius are all standard. The net force shall always point towards the center. In this case the net force points downwards.

What are the forces on this object?

There are two forces on the object at this moment:

Weight, $W$ , which points downwards. $W = m\cdot g = 0.20\times 9.81 = 1.962\;\text{N}$ .
Tension, $T$ , which also points downwards. The size of the tension force needs to be found.

What's the size of the tension force?

Gravity and tension points in the same direction. The size of their resultant force is the sum of the two forces. In other words,

$\Sigma F = T + W$ .

$T = \Sigma F - W = 5.0625 - 1.962 = 3.1$ .

All three values in this question are given with two sig. fig. Round the value of $T$ to the same number of significant figures.

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2 years ago