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pressure on gas & liquids


apurvmj
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Hi all,

It is said that liquids are mostly non compressible & gases are compressible.

Could some one pls explain, do both require same energy to have same pressure?

For example how much energy it requires to compress air of 1 m3 to 10 kg/cm2 & water of 1 m3 to same10 kg/cm2

Thanks.

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For example how much energy it requires to compress air of 1 m3 to 10 kg/cm2 & water of 1 m3 to same10 kg/cm2

Thanks.

 

Simply

 

Energy = Force * Length-->(Force is vector)

 

Your asking is ???

kg/cm2 ---> kgf/cm2 (kgf = 9.8 kgm m/s2)

 

Pressure.

Pressure = Force /Area

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The energy is going to be given by [math]\Delta{P}\Delta{V}[/math], and there must be assumptions that this happens with no change in temperature and no heat flow. It also depends on how you get from the original values to the final ones — the work is path dependent.

 

http://en.wikipedia.org/wiki/Work_(thermodynamics)#Pressure-volume_work

 

The amount by which something compresses under a uniform pressure is the bulk modulus

http://en.wikipedia.org/wiki/Bulk_modulus

 

For water, this is ~2.2×10^9 Pa, while for air it's around 10^5 Pa. So air compresses a lot more for a given pressure increase. Or, conversely, it takes a much larger pressure difference for water to achieve the same change in volume as air.

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correct me if I'm wrong

for my 'example' part,

I got one formula for gases

energy= pressure*volume.

Is same applicable for liquids?

I'm keen to know that does water & air of same volume & same pressure hold the same energy?

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correct me if I'm wrong

for my 'example' part,

I got one formula for gases

energy= pressure*volume.

Is same applicable for liquids?

I'm keen to know that does water & air of same volume & same pressure hold the same energy?

 

The same holds for liquids, but they are often treated as completely incompressible, in which case the energy depends on the pressure alone.

 

I don't think you can compare same volume and pressure because the materials respond differently to a change in pressure. Whatever you get will only be true at one set of values, and the equations are used to find the energy released/absorbed when you change the values.

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But I'm curious to compare that for different kind of materials, keeping some parameters same like volume & pressure, how much energy they hold.

I can find it for air as pressure * volume = 106 * 1 = 1000000 J (10 kg/cm2 = 106 N/m2)

But how can I find the energy hold by water of 1 m3 at 10 kg/cm2 .

Edited by apurvmj
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Energy = Force * Length-->(Force is vector)

Energy is not defined that way. That expression merely the work done on a system. Work is not a form of energy. It merely has the same dimensions of energy.

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Energy is not defined that way. That expression merely the work done on a system. Work is not a form of energy. It merely has the same dimensions of energy.

 

Energy

 

Dimension

ML2T -2

 

M; mass

L; length

T; time

 

Unit

kgm2/s2

dyne.cm, N.m , erg, J, Cal, Btu

 

For dimension calculation

E=F.L

P.V= (F/A).V=F.L

 

Work and heat are transferred between system and surround.

They are not state functions, but are path functions.

 

Real energy(work) is calculated by using path integration function.

Edited by alpha2cen
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You must distinguish internal energy from enthalpy to make a sense of the question.

 

Take 1m3 of an idealized liquid with zero compressibility: if it's in a tank and you push on it to increase the pressure by 1Pa, as your piston doesn't move, it takes zero joule of work - here it's internal energy. But if you push this liquid from one tank to an other tank at 1Pa higher pressure, you piston moves and consumes 1J of mechanical work - here it's enthalpy.

 

One subtlety of enthalpy is that it is not stored fully in the fluid itself; part is energy stored in the surroundings of the fluid (for instance the gas that top the liquid in the tanks at pressures P and P+1Pa). Despite of this, enthalpy is a property of this fluid alone, depending just on the state (pressure, density...) of this fluid, and you need no information about the surroundings (composition, heat capacity...) to compute it. Abstract, isn't it? And very powerful theory.

 

It also means that work does not correspond uniquely to the state of the fluid; in some processes it may correspond nearly to enthalpy changes, in others to changes in ambiant heat... This allows thermal engines to extract a net mechanical work from a fluid making a closed cycle, for instance get more work from an expansion at high temperature than is invested in the compression at low temperature.

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