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kirza4 [7]
3 years ago
15

A mass of gas has a volume of 4m3, a temperature of 290k, and an absolute pressure of 475 kpa. When the gas is allowed to expand

to 6.5m3, it's new temperature is 277K. What's the absolute pressure of the gas after expansion?
Physics
2 answers:
bogdanovich [222]3 years ago
5 0
Liquid alcohol will change to vapor when its temperature teaches the point
motikmotik3 years ago
4 0

Given:

V1 = 4m3

T1 = 290k

P1 = 475 kpa = 475000 Pa

V2 = 6.5m3

T2 = 277K

Required:

P

Solution:

n = PV/RT

n = (475000 Pa)(4m3) / (8.314 Pa-m3/mol-K)(290k)

n = 788 moles

P = nRT/V

P  = (788 moles)(8.314 Pa-m3/mol-K)(277K)/(6.5m3)

P = 279,204 Pa or 279 kPa

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Answer:

a)  k_{e} = 928 J , b)U = -62.7 J , c) K = 0 , d) Y = 11.0367 m,  e)  v = 15.23 m / s  

Explanation:

To solve this exercise we will use the concepts of mechanical energy.

a) The elastic potential energy is

      k_{e} = ½ k x²

      k_{e} = ½ 2900 0.80²

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b) place the origin at the point of the uncompressed spring, the spider's potential energy

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     U = 8 9.8 (-0.80)

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c) Before releasing the spring the spider is still, so its true speed and therefore the kinetic energy also

      K = ½ m v²

      K = 0

d) write the energy at two points, maximum compression and maximum height

     Em₀ = ke = ½ m x²

     E_{mf} = mg y

     Emo = E_{mf}

     ½ k x² = m g y

     y = ½ k x² / m g

     y = ½ 2900 0.8² / (8 9.8)

     y = 11.8367 m

As zero was placed for the spring without stretching the height from that reference is

     Y = y- 0.80

     Y = 11.8367 -0.80

     Y = 11.0367 m

Bonus

Energy for maximum compression and uncompressed spring

     Emo = ½ k x² = 928 J

     E_{mf}= ½ m v²

     Emo = E_{mf}

     Emo = ½ m v²

      v =√ 2Emo / m

     v = √ (2 928/8)

     v = 15.23 m / s

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Answer:

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