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castortr0y [4]
3 years ago
6

Suppose 3 mol of neon (an ideal monatomic gas) at STP are compressed slowly and isothermally to 0.19 the original volume. The ga

s is then allowed to expand quickly and adiabatically back to its original volume.
Required:
a. Find the highest temperature attained by the gas.
b. Find the lowest temperature attained by the gas.
c. Find the highest pressure attained by the gas.
d. Find the lowest pressure attained by the gas.
Physics
1 answer:
Radda [10]3 years ago
6 0

Answer:

a. 273 K b. 90.1 K c. 5.26 atm d. 0.33 atm

Explanation:

For isothermal expansion PV = constant

So, P₁V₁ = P₂V₂ where P₁ = initial pressure of gas = 1 atm (standard pressure), V₁ = initial volume of gas, P₂ = final pressure of gas and V₂ = final volume of gas,

So, P₁V₁ = P₂V₂

P₂ = P₁V₁/V₂

Since V₂/V₁ = 0.19,

P₂ = P₁V₁/V₂

P₂ = 1 atm (1/0.19)  

P₂ = 5.26 atm

For an adiabatic expansion, PVⁿ = constant where n = ratio of molar heat capacities = 5/3 for monoatomic gas

So, P₂V₂ⁿ = P₃V₃ⁿ where P₂ = initial pressure of gas = 5.26 atm, V₂ = initial volume of gas, P₃ = final pressure of gas and V₃ = final volume of gas,

So, P₂V₂ⁿ = P₃V₃ⁿ

P₃ = P₂V₂ⁿ/V₃ⁿ

P₃ = P₂(V₂/V₃)ⁿ

Since V₃ = V₁ ,V₂/V₃ = V₂/V₁ = 0.19

1/0.19,

P₃ = P₂(V₂/V₃)ⁿ

P₃ = 5.26 atm (0.19)⁽⁵/³⁾

P₃ = 5.26 atm × 0.0628

P₃ = 0.33 atm

Using the ideal gas equation

P₃V₃/T₃ = P₄V₄/T₄ where P₃ = pressure after adiabatic expansion = 0.33 atm , V₃ = volume after adiabatic expansion, T₃ = temperature after adiabatic expansion  P₄ = initial pressure of gas = P₁ = 1 atm , V₄ = initial volume of gas = V₁ and T₄ = initial temperature of gas = T₁ = 273 K (standard temperature)

P₃V₃/T₃ = P₄V₄/T₄

T₃ = P₃V₃T₄/P₄V₄    

T₃ = (P₃/P₄)(V₃/V₄)T₂

Since V₃ = V₄ = V₁ and P₄ = P₁

V₃/V₄ = 1 and P₃/P₄ = P₃/P₁

T₃ = (P₃/P₁)(V₃/V₄)T₂

T₃ = (0.33 atm/1 atm)(1)273 K  

T₃ = 90.1 K

So,

a. The highest temperature attained by the gas is T₁ = 273 K

b. The lowest temperature attained by the gas = T₃ = 90.1 K

c. The highest pressure attained by the gas is P₂ = 5.26 atm

d. The lowest pressure attained by the gas is P₃ = 0.33 atm

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The change in potential energy when the block falls to ground is -480J.

The maximum change in kinetic energy of the ball is 480 J.

The initial kinetic energy of the ball is 0 J.

The final  kinetic energy of the ball is 0.148J.

The initial potential energy of the ball is 0.187 J.

The final  potential energy of the ball is 0 J.

The work done by the air resistance is 0.039 J.

<h3>Change in potential energy when the block falls to ground</h3>

ΔP.E = -mgh

ΔP.E = -Wh

ΔP.E = - 40 x 12

ΔP.E = -480 J

<h3>Maximum change in kinetic energy of the ball</h3>

ΔK.E = - ΔP.E

ΔK.E = - (-480 J)

ΔK.E = 480 J

<h3>Initial kinetic energy of the ball</h3>

K.Ei = 0.5mv²

where;

  • v is zero since it is initially at rest

K.Ei = 0.5m(0) = 0

<h3>Final kinetic energy</h3>

K.Ef =  0.5mv²

K.Ef = 0.5(0.0091)(5.7)²

K.Ef = 0.148 J

<h3>Initial potential energy of the ball</h3>

P.Ei = mghi

P.Ei = 0.0091 x 9.8 x 2.1

P.Ei = 0.187 J

<h3>Final potential energy</h3>

P.Ef = mghf

P.Ef = 0.0091 x 9.8 x 0

P.Ef = 0

<h3>Work done by the air resistance</h3>

W = ΔE

W = P.E - K.E

W = 0.187 J - 0.148 J

W = 0.039 J

Learn more about potential energy here: brainly.com/question/1242059

#SPJ1

<h3 />
7 0
1 year ago
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