Why does the pressure change on uniformly mixing two liquids? - Physics Stack Exchange most recent 30 from physics.stackexchange.com 2019-08-21T18:03:53Z https://physics.stackexchange.com/feeds/question/450944 http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://physics.stackexchange.com/q/450944 3 Why does the pressure change on uniformly mixing two liquids? user218336 https://physics.stackexchange.com/users/0 2018-12-29T07:34:16Z 2018-12-29T14:46:31Z <p><a href="https://i.stack.imgur.com/McTra.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/McTra.png" alt="enter image description here"></a></p> <p>The two cylinders are connected the upper cylinder has a cross-section of A and the lower one has a cross-section of 2A (I've taken the cross-sections to be A and 2A as they are easier to work with than the radii r and 2r). </p> <p>Initially the liquid present in the upper cylinder has a density of <span class="math-container">$\rho$</span> and the lower one has a density of <span class="math-container">$2\rho$</span>. The pressure at the bottom in this scenario can be calculated as: </p> <p><span class="math-container">$$P_1=\rho\cdot g\cdot 2h+2\rho\cdot g \cdot h=4\rho\cdot g\cdot h$$</span> </p> <p>Now, if we were to mix the two liquids present in the cylinders to create a new solution with a uniform density of <span class="math-container">$\frac32\rho$</span> (The volume of both the cylinders is the same so we can simply take the mean of the two densities). </p> <p>The pressure in this case can be calculated as:<br> <span class="math-container">$$P_2=\frac32\rho\cdot g \cdot 3h= \frac92\cdot g \cdot h$$</span> </p> <p>This was one approach to calculate the pressures another method is by simply calculating <span class="math-container">$$\frac{F}{A}$$</span> </p> <p>Since the entire liquid system is in equilibrium and the only external force balancing the gravitational force is the normal applied at the bottom surface the pressure will be: </p> <p><span class="math-container">$$P=\frac{Mg}{A}$$</span><br> <span class="math-container">$$P=\frac{\rho\cdot 2h \cdot A\cdot g + 2\rho\cdot 2A \cdot h \cdot g}{2\cdot A }= 3\rho \cdot g \cdot h$$</span></p> <p>Which is matching with neither of the cases but i can't see why this is wrong.</p> https://physics.stackexchange.com/questions/450944/-/450974#450974 1 Answer by Deep for Why does the pressure change on uniformly mixing two liquids? Deep https://physics.stackexchange.com/users/122958 2018-12-29T12:16:50Z 2018-12-29T12:16:50Z <p>First consider the case of a cylinder of uniform cross section <span class="math-container">$A$</span>, in which liquids of density <span class="math-container">$\rho_1,\rho_2,$</span> with corresponding volumes <span class="math-container">$V_1,V_2,$</span> are present. Using the hydrostatic equation, the pressure at the bottom is <span class="math-container">$P_1=\rho_1gh_1+\rho_2gh_2=\rho_1g(V_1/A)+\rho_2g(V_2/A)=(\rho_1V_1+\rho_2V_2)g/A$</span>. If after mixing the total volume doesn't change then mass conservation says that the final density should be <span class="math-container">$\rho_f=(\rho_1V_1+\rho_2V_2)/(V_1+V_2)$</span>. The corresponding pressure at the bottom will be: <span class="math-container">$\rho_fgh_f=\rho_fg(h_1+h_2)=\rho_fg(V_1+V_2)/A=(\rho_1V_1+\rho_2V_2)g/A,$</span> the same as before. Therefore, when the cylinder has uniform cross-section, mixing doesn't change the pressure at the bottom, which is logical since the same weight of fluid is supported by the bottom in both cases.</p> <p>In your problem the cross-section varies. Let <span class="math-container">$\rho_1,V_1,A_1$</span> and <span class="math-container">$\rho_2,V_2,A_2$</span> be the density, volume and cross-section of the two fluids. The interface between the two fluids lies exactly where the cross-section changes. Using the hydrostatic equation, the pressure is: <span class="math-container">$P_1=\rho_1gh_1+\rho_2gh_2=\rho_1g(V_1/A_1)+\rho_2g(V_2/A_2)$</span>, which can't be simplified further. After mixing the density becomes <span class="math-container">$\rho_f=(\rho_1V_1+\rho_2V_2)/(V_1+V_2)$</span> as before. But now the pressure at the bottom changes to <span class="math-container">$\rho_fgh_f\neq P_1,$</span> as you can verify by substitution. Since the same weight of fluid is being supported before and after mixing, is the change in pressure a contradiction? No, it isn't.</p> <p>Above result derived using hydrostatic equation is correct. If you want to derive the pressure on the bottom using force balance then you must recognise that in the case where cross-section of the container varies, weight of the fluid is not the only force acting on the bottom. There is also a downward reaction force due to the horizontal wall where the cross-section changes (denoted by black arrows in the figure below). This reaction force is equal to the product of hydrostatic pressure at the horizontal wall (equal to <span class="math-container">$\rho_1gh_1$</span> before mixing and <span class="math-container">$\rho_fgh_1$</span> after mixing) and its area. When this is accounted for you get the same answer as above.</p> <p>In short, the change in pressure at the bottom subsequent to mixing is entirely due to a change in the hydrostatic pressure at the horizontal wall.</p> <p><a href="https://i.stack.imgur.com/lZ6Uym.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lZ6Uym.png" alt="test"></a></p>