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Lorico [155]
3 years ago
10

An ideal monatomic gas initially has a temperature of T and a pressure of p. It is to expand from volume V1 to volume V2. If the

expansion is isothermal, what are the final pressure pfi and the work Wi done by the gas? If, instead, the expansion is adiabatic, what are the final pressure pfa and the work Wa done by the gas? State your answers in terms of the given variables.
Physics
1 answer:
yawa3891 [41]3 years ago
3 0

Answer:

Isothermal :   P2 = ( P1V1 / V2 ) ,  work-done pdv = nRT * In( \frac{V2}{v1} )

Adiabatic : : P2 = \frac{P1V1^{\frac{5}{3} } }{V2^{\frac{5}{3} } }  , work-done =

W = (3/2)nR(T1V1^(2/3)/(V2^(2/3)) - T1)

Explanation:

initial temperature : T

Pressure : P

initial volume : V1

Final volume : V2

A) If expansion was isothermal calculate final pressure and work-done

we use the gas laws

= PIVI = P2V2

Hence : P2 = ( P1V1 / V2 )

work-done :

pdv = nRT * In( \frac{V2}{v1} )

B) If the expansion was Adiabatic show the Final pressure and work-done

final pressure

P1V1^y = P2V2^y

where y = 5/3

hence : P2 = \frac{P1V1^{\frac{5}{3} } }{V2^{\frac{5}{3} } }

Work-done

W = (3/2)nR(T1V1^(2/3)/(V2^(2/3)) - T1)

Where    T2 = T1V1^(2/3)/V2^(2/3)

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(11%) Problem 5: A uniform stationary ladder of length L = 2.7 m and mass M = 11 kg leans against a smooth vertical wall, while
jeka94

Answer:

970.2 N

Explanation:

We are given that

Length of ladder=2.7 m

Mass,M=11 kg

Coefficient of friction=\mu=0.45

\theta=51^{\circ}

Mass of painter=8M

Distance from base=d

We have to find the magnitude of the normal force exerted by the floor on the ladder.

Normal force exerted by floor on the ladder=mg+8mg=9mg

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7 0
3 years ago
When an element ejects an alpha particle, the mass number of that element
Katyanochek1 [597]

Answer:

A.reduce by 4.

Explanation:

In Chemistry, an alpha particle which is typically a helium nucleus has an atomic mass (number of nucleons) of four (4). Therefore, the ejection of an alpha particle by any chemical element simply means that, the mass number of that particular chemical element would be reduced by four (4).

Basically, this process is referred to as an alpha decay and can be defined as a radioactive disintegration of a chemical element that causes it to emit an alpha particle and as a result reducing its mass number by four (4).

Hence, when an element ejects an alpha particle, the mass number of that element reduces by 4.

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The tires of a car support the weight of a stationary car. If one tire has a slow leak, the air pressure within the tire will __
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Answer:

The tires of a car support the weight of a stationary car. If one tire has a slow leak, the air pressure within the tire will decrease with time, the surface area between the tire and the road will increase with time, and the net force the tire exerts on the road will be constant with time.

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when a wheel has an air leak, it means that the inside of the tire has less air, which means that there will be less air pushing the walls of the tire so that the air pressure decreases.

On the other hand, the tire begins to deform due to lack of air which increases the area of ​​contact with the floor.

As the weight of the car remains constant and the air has a negligible mass the force towards the road will be the same

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4 years ago
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