MHB How to compute the energy needed to compress the water isothermally?

  • Thread starter Thread starter WMDhamnekar
  • Start date Start date
  • Tags Tags
    Energy Water
AI Thread Summary
The discussion centers on calculating the energy required to compress water isothermally, with an initial answer of 29.4 Joules being questioned. The expert's calculations suggest that the energy needed is significantly higher, around 506 Joules, based on constant pressure assumptions. However, it is noted that pressure builds from 1 to 100 atmospheres, indicating the actual energy requirement would be approximately half of that estimate, around 229 Joules. The temperature of the water at 20°C is relevant, as the isothermal compressibility coefficient is acknowledged to align with the calculations. Ultimately, the initial answer of 29.4 Joules appears to be incorrect based on the provided analysis.
WMDhamnekar
MHB
Messages
376
Reaction score
28
Hi,
Answer given is $E_n=29.4 Joules$ Here is the question.

1602942344841.png


Answer provided by the Chemistry math expert/Professor is as follows but it is different from the answer given. How is that?

Compressibility is the fractional change in volume per unit increase in pressure. For each atmosphere increase in pressure, the volume of water would decrease 46.4 parts per million.
I'll pick a shape for the device, calculate distance traveled and force required, and use $work = force \times distance.$
with 100 atm, volume would decrease by 4640 PPM or by a factor of 0.00464

10 kg of water is about 10 liters or $0.01 m^3$

$100 atm = 1.013e7 Pa$ or $1.013e7 N/m^2$
assume a cube shape, height is $\sqrt[3]{0.01 m^3} = 0.2154435 m$ and base area is$ 0.0464159 m^2$ (coincidence that "464" appears as two different values)
force on piston is $1.013e7 N/m^2 \times 0.0464 m^2 = 470000 N$
that change in volume causes what change in height

new volume $= 0.01 m^3 – 0.01 m^3(0.00464) = 0.00995 m^3$

which has a height of $\frac{0.00995 m^3}{0.0464159 m^2} = 0.2143663 ,$

change is 0.2154435 – 0.2143663 = 0.00108 m
energy = Fd = (470000 N)(0.00108 m) = 506 JIs temperature of water $20^\circ C$ to be considered?
 
Last edited:
Mathematics news on Phys.org
Dhamnekar Winod said:
Answer given is $E_n=29.4 Joules$ Here is the question.

Answer provided by the Chemistry math expert/Professor is as follows but it is different from the answer given. How is that?
(snip)
force on piston is $1.013e7 N/m^2 \times 0.0464 m^2 = 470000 N$
that change in volume causes what change in height
new volume $= 0.01 m^3 – 0.01 m^3(0.00464) = 0.00995 m^3$
which has a height of $\frac{0.00995 m^3}{0.0464159 m^2} = 0.2143663 ,$
change is 0.2154435 – 0.2143663 = 0.00108 m

energy = Fd = (470000 N)(0.00108 m) = 506 J
Your expert's answer assumes that the force/pressure is constant at 100 atmosphere, but that is not the case.
Instead it will build up from 1 atmosphere up to 100 atmosphere.
So we can expect the actual answer to be about half of that $506\,J$, which is really an upper estimate.

I found $229\,J$ myself while taking the changing pressure into account with Calculus, which is indeed in the neighborhood of half of that $506\,J$.
Either way, it looks as if the answer of $29.4\,J$ is not correct.

Is temperature of water $20^\circ C$ to be considered?

We take the coefficient of isothermal compressibility of water at $20^\circ C$.
Wikipedia mentions that it is $4.4$ to $5.1×10^{-10}\, Pa^{-1}$ in ordinary conditions.
Close enough to that 46.4 parts per million that you mentioned.
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Back
Top