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Georgia [21]
3 years ago
10

Three kilograms of argon (Ar) changes from an initial volume and a temperature of 298K to (a) four times the volume and a temper

ature of 298K and (b) one fourth the original volume under adiabatic conditions. In each case: Calculate the amount of work performed. Is the work done by the system or the surroundings? (pl/, for monatomic gases is 1.667)
Physics
1 answer:
Kay [80]3 years ago
8 0

Answer:

Explanation:

3 kg = 3 / 40 = .075 moles = n

a ) Since the gas is expanding isothermally ( temperature being constant )

work done by the gas

= 2.303 n RT log V₂ / V₁

Here V₂ / V₁ = 4 , T = 298

Put these values in the equation above ,

work done = .075x 2.303 x 8.312 x 298 log 4

= 257.6 J

b) In adibatic change

pv^\gamma = constant

T V^{\gamma-1} = constant

T₁ / T₂ = (\frac{V_2}{V_1}  )^{\gamma-1}

298 / T₂ = ( 4)^{1.667-1}

T₂ = 751.26 K.

In adiabatic change work done

= n R ( T₁ - T₂) / (γ -1)

.075x 8.312 X ( 751 - 298 ) / .667

= 423.38. J

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Answer:

The answer to your question is:

a) t = 3.81 s

b) vf =  37.4 m/s

Explanation:

Data

height = 71.3 m = 234 feet

t = 0 m/s

vf = ?

vo = 0 m/s

Formula

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vf = vo + gt

Process

a)

               h = vot + 1/2gt²

             71.3 = 0t + 1/2(9.81)t²

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How many cubic feet are there in a gallon?
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A small box of mass m1 is sitting on a board of mass m2 and length L. The board rests on a frictionless horizontal surface. The
Nadusha1986 [10]

Answer:

The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is \left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}

Explanation:

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\sum {F = ma}

Here, is the sum of all the forces on the object, mm is mass of the object, and aa is the acceleration of the object.

The expression for static friction over a horizontal surface is,

F_{\rm{f}}} \leq {\mu _{\rm{s}}}mg

Here, {\mu _{\rm{s}}} is the coefficient of static friction, mm is mass of the object, and g is the acceleration due to gravity.

Use the expression of static friction and solve for maximum static friction for box of mass {m_1}

Substitute  for in the expression of maximum static friction {F_{\rm{f}}} = {\mu _{\rm{s}}}mg

{F_{\rm{f}}} = {\mu _{\rm{s}}}{m_1}g

Use the Newton’s second law for small box and solve for minimum acceleration aa to pull the box out.

Substitute  for , [/tex]{m_1}[/tex] for in the equation .

{F_{\rm{f}}} = {m_1}a

Substitute {\mu _{\rm{s}}}{m_1}g for {F_{\rm{f}}} in the equation {F_{\rm{f}}} = {m_1}a

{\mu _{\rm{s}}}{m_1}g = {m_1}a

Rearrange for a.

a = {\mu _{\rm{s}}}g

The minimum acceleration of the system of two masses at which box starts sliding can be calculated by equating the pseudo force on the mass with the maximum static friction force.

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Use the Newton’s second law for the system of box and the board.

Substitute for for in the equation .

{F_{\min }} = \left( {{m_1} + {m_2}} \right)a

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There is no friction between the board and the surface. So, the force required to accelerate the system with the minimum acceleration to slide the box over the board is equal to total mass of the board and box multiplied by the acceleration of the system.

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