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Lorico [155]
2 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]2 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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A hydraulic lift raises a 2000 kg automobile when a 500 N force is applied to the smaller piston. If the smaller piston has an a
Inga [223]

Answer:

The cross-sectional area of the larger piston is 392cm ^{2}[/tex]

Explanation:

To solve this problem we apply the following formula:

Pascal principle: F=P*A   Formula (1)

F=Force applied to the piston

P: Pressure

A= Piston area

Nomenclature:

Fp= Force on the primary piston= 500N

W= car weight =m*g=2000kg*9.8m/s2= 19600N

Fs= Force on the secondary piston= W = 19600N

Ap= Primary piston area=10cm^{2} =10*10^{-4}m^{2}

As= Secondary piston area=?

The pressure acting on one side is transmitted to all the molecules of the liquid because the liquid is incompressible.

In the equation (1)

P=F/A

Pp=Ps

\frac{Fp}{Ap} = \frac{Fs}{As}

As= \frac{Fs*Ap}{Fp}

As=\frac{19600*10*10^{-4} }{500}

As=0.0392m^{2} =0.0392*10^{4}cm^{2}

As=392cm ^{2}

5 0
2 years ago
A forklift does 450 J of work to lift a 150 N hot tub. How high is the hot tub lifted? Round your answer to the nearest whole nu
vivado [14]
There are already 2 givens, hence we can do direct substitution to get the answer. To make the process simpler, derive the distance formula from the Work formula.

Work = Force x Distance

Distance = \frac{Work}{Force}


Work is 450J while the force is 150N hot tub

To get the proper units, get the equivalent of Joule to eliminate newton. A joule Is equal to 1 N-m

Distance = \frac{450N-m}{150N}
Distance = 3m

Hence, the hot tub is lifted 3 meters.
5 0
2 years ago
Read 2 more answers
All of these are examples of pseudopsychology EXCEPT:
jeyben [28]

Answer:

psychoanalysis

Explanation:

psychoanalysis a system of psychological theory and therapy which aims to treat mental disorders by investigating the interaction of conscious and unconscious elements in the mind and bringing repressed fears and conflicts into the conscious mind by techniques such as dream interpretation and free association.

4 0
2 years ago
Read 2 more answers
En el País Vasco los deportistas rurales levantan enormes piedras hasta su hombro. En un concurso , Jose levanta una piedra de 2
Mama L [17]

Answer:

Txomin lifted the stone with greater mass. (Txomin levantó la roca con mayor fuerza).

Explanation:

The sportsman that lifts the stone with a greater mass needs a higher force (El deportista que levanta la piedra con mayor masa necesita una mayor fuerza):

José

F = (200\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)

F = 1961.4\,N

Txomin

F = (220\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)

F = 2157.54\,N

Txomin lifted the stone with greater mass. (Txomin levantó la roca con mayor fuerza).

5 0
1 year ago
A pilot in a small plane encounters shifting winds. He flies 26.0 km northeast, then 45.0 km due north. From this point, he flie
cluponka [151]

Answer:

a) v₃ = 19.54 km, b)  70.2º north-west

Explanation:

This is a vector exercise, the best way to solve it is finding the components of each vector and doing the addition

vector 1 moves 26 km northeast

let's use trigonometry to find its components

         cos 45 = x₁ / V₁

         sin 45 = y₁ / V₁

         x₁ = v₁ cos 45

         y₁ = v₁ sin 45

         x₁ = 26 cos 45

         y₁ = 26 sin 45

         x₁ = 18.38 km

         y₁ = 18.38 km

Vector 2 moves 45 km north

        y₂ = 45 km

Unknown 3 vector

          x3 =?

          y3 =?

Vector Resulting 70 km north of the starting point

           R_y = 70 km

we make the sum on each axis

X axis

      Rₓ = x₁ + x₃

       x₃ = Rₓ -x₁

       x₃ = 0 - 18.38

       x₃ = -18.38 km

Y Axis

      R_y = y₁ + y₂ + y₃

       y₃ = R_y - y₁ -y₂

       y₃ = 70 -18.38 - 45

       y₃ = 6.62 km

the vector of the third leg of the journey is

         v₃ = (-18.38 i ^ +6.62 j^ ) km

let's use the Pythagorean theorem to find the length

         v₃ = √ (18.38² + 6.62²)

         v₃ = 19.54 km

to find the angle let's use trigonometry

           tan θ = y₃ / x₃

           θ = tan⁻¹ (y₃ / x₃)

           θ = tan⁻¹ (6.62 / (- 18.38))

           θ = -19.8º

with respect to the x axis, if we measure this angle from the positive side of the x axis it is

          θ’= 180 -19.8

          θ’= 160.19º

I mean the address is

          θ’’ = 90-19.8

          θ = 70.2º

70.2º north-west

3 0
2 years ago
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