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3 years ago

0.0884 moles of a diatomic gas

Physics

2 answers:

Sloan [31]3 years ago

7 0

Answer:

W = - 118.24 J (negative sign shows that work is done on piston)

Explanation:

First, we find the change in internal energy of the diatomic gas by using the following formula:

$\Delta\ U = nC_{v}\Delta\ T$

where,

ΔU = Change in internal energy of gas = ?

n = no. of moles of gas = 0.0884 mole

Cv = Molar Specific Heat at constant volume = 5R/2 (for diatomic gases)

Cv = 5(8.314 J/mol.K)/2 = 20.785 J/mol.K

ΔT = Rise in Temperature = 18.8 K

Therefore,

$\Delta\ U = (0.0884\ moles)(20.785\ J/mol.K)(18.8\ K)\\\Delta\ U = 34.54\ J$

Now, we can apply First Law of Thermodynamics as follows:

$\Delta\ Q = \Delta\ U + W$

where,

ΔQ = Heat flow = - 83.7 J (negative sign due to outflow)

W = Work done = ?

Therefore,

$-83.7\ J = 34.54\ J + W\\W = -83.7\ J - 34.54\ J\\$

W = - 118.24 J (negative sign shows that work is done on piston)

Mandarinka [93]3 years ago

7 0

Answer:

-49.2

Explanation:

Trust me bro

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Answer:

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3 years ago

According to Newton’s law of universal gravitation, which statements are true?

andreyandreev [35.5K]

Before we solve this, we should know this fact:

According to Newton's Law of Gravitation, the force between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. The force acts along the line joining the centres of the two objects. It can be shown by this:

$F ∝ \frac{Mm}{ {d}^{2} }$

Now, let us check all the options.

A. As we move to higher altitudes, the force of gravity on us decreases.

This statement is true.

The force of gravity is inversely proportional to the square of distance from the centre of the earth. If, we go up the surface of the earth, the distance from the centre of the earth increases and hence the value of force of gravity decrease. So, force of gravity decreases with altitude.

B. As we move to higher altitudes, the force of gravity on us increases.

This statement is false.

We have already got the result in option A. that the force of gravity decreases with altitude. It never increases with altitude.

C. As we gain mass, the force of gravity on us decreases.

This statement is false.

The force of gravity is directly proportional to the product of the masses. So, if increase our mass, then the force of gravity will also increase and if we decrease our mass, then the force of gravity decreases.

D. As we gain mass, the force of gravity on us increases.

This statement is true.

As mentioned earlier in option C., the force of gravity is directly proportional to the product of the masses of the earth and another object. So, as we gain mass, the force of gravity on us increases.

E. As we move faster, the force of gravity on us increases.

This statement is true.

Here, we have to consider a different formula. According to Newton's Second Law,

F = ma, where F is the force, m is the mass and a is the acceleration.

In other words,

F ∝ a, i.e., force is directly proportional to acceleration.

We know, acceleration is the rate of change of velocity of an body within a time period.

So, if speed is increased, then acceleration will also be greater, which results in the increase of force. So, as we move faster, the force of gravity on us increases.

Answers:

A: As we move to higher altitudes, the force of gravity on us decreases.

D: As we gain mass, the force of gravity on us increases.

E: As we move faster, the force of gravity on us increases.

Hope you could understand.

If you have any query, feel free to ask.

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2 years ago

A satellite orbits earth with a mean altitude of 361 km. If the orbit is circular, what are the satellite's time period and spee

Advocard [28]

Answer:

v = 7.69 x 10³ m/s = 7690 m/s

T = 5500 s = 91.67 min = 1.53 h

Explanation:

In order for the satellite to orbit the earth, the force of gravitation on satellite must be equal to the centripetal force acting on it:

$F_{gravitation}= F_{centripetal}\\\\\frac{GM_{s} M_{E}}{r^2} = \frac{M_{s} v^2}{r}\\\\\frac{GM_{E}}{r} = v^2\\\\v = \sqrt{\frac{GM_{E}}{r} } \\\\$

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G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

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v = orbital speed = ?

Therefore,

$v = \sqrt{\frac{(6.67 x 10^{-11}N.m^2/kg^2)(5.97 x 10^{24} kg)}{6.732 x 10^6 m} }\\\\$

v = 7.69 x 10³ m/s

For time period satellite completes one revolution around the earth. It means that the distance covered by satellite is equal to circumference of circle at the given altitude.

So, its orbital speed can be given as:

$v = \frac{Circumference of Circle at Given Altitude}{T}\\\\v = \frac{2\pi r}{T}\\\\$

where,

T = Time Period of Satellite = ?

Therefore,

$T = \frac{2\pi r}{v}\\\\T = \frac{(2\pi )(6.732 x 10^6 m}{7.69 x 10^3 m/s}\\\\$

T = 5500 s = 91.67 min = 1.53 h

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