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If you cannot understand, you are not alone; even teachers find difficulty to grab. The difficulty of understanding is due to the rubbish ways of explanation in academic/ textbook. Let's see the actual concept. Cv is heat capacity at constant volume. Constant volume has no direct meaning in heat retaining capacity. Constant volume indirectly means work ...

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Volume is not a meaningful measure of quantity, for the reason you hint at in your question. You can say how many moles (or grams) of water you drank - more useful if you want to know about the impact on your body chemistry. This is related to my answer about scales measuring in grams rather than Newtons. Can you see how?

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When ice is heated from 0 to 4 degrees C, it actually contracts. The water molecules get closer together and the water occupies less volume. However, above 4 degrees C water expands(i.e. in your case). This is explained by Charles's law but any way this change in volume is not very drastic. This calculator might be helpful

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$1$ litre of water will remain almost $1$ litre as long as it is in the liquid state, no matter what the temperature is. The following formula gives you an order of magnitude estimate of the expansion: $$\Delta V=V_0\ \Delta T \ \beta$$ where $\beta$ is the coefficient of thermal expansion and $V_0$ is the initial volume. For water, \$\beta \approx 10^{-...

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You need some flowmeters. The type depends on the quantity of the gas that you use, and the type. You would then need to come up with a full description of a program to use those values. For example, you might just want to store in a file the quantity of gas supplied to furnaces as a function of time, updated every second (or minute). Or you might want to ...

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You already have the answer when you write $$\frac{\Delta h}{h} = (\beta -2\alpha)\Delta T$$ What you do after that is unnecessary and does not make sense. You have already said that the height of the tube is irrelevant, so the height of the liquid "relative to the tube" is meaningless. If initially the liquid fills the tube completely and you want to ...

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Assuming there is only one molecular in this box and assuming it is a closed system with initial P-V state is defined. The question becomes: can the system move to anywhere on the PV diagram? Well we can adjust volume to any number. Then the question becomes: can pressure reaches to any values on PV diagram? Pressure relates to impacting intensity and ...

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In a polytropic process other than adiabatic, you are controlling the temperature in tandem with P and V in such a way that n is constant. You can certainly achieve negative values of n by controlling the temperature appropriately. From the ideal gas law, if T and P are expressed parametrically in terms of V, then:$$\frac{P}{P_0}=\left(\frac{V_0}{V}\right)^... 3 It's simply inherent to the definition of polytropic processes that they don't allow the system to increase both its pressure and volume at the same time. That doesn't mean you can't increase a system's pressure and volume. You just need a non-polytropic process to do so. For example, it could be a compound process consisting of two polytropic processes with ... 3 P-V work is not the only kind of work that can be done on the contents of your system. In the case of your fan example, the fan is doing work on the gas within the container by exerting force on it through a displacement (of the fan blade). The kinetic energy imparted to the gas by the fan is then converted to internal energy by viscous dissipation (a ... 0 OK, I can not show the math, but anything you try to fit inside the box, will itself be curved/stretched/contracted along with the space. Therefore, you should not be able to fit more stuff inside one box as compared to the other of the same size. Even saying same size involves space and its curving. So, even your box will be curved as there is nothing that ... 2 Within the Schwarzschild metric, the volume does change. It is the rectangle formed by the radial dimension and time which is invariant: The dilating effect of the Schwarzschild metric$$ \mathrm ds^2 = -\left(1 - \frac{2GM}{c^2 r}\right) c^2 ~\mathrm dt^2 + \frac{1}{1 - \frac{2GM}{c^2 r} }~\mathrm dr^2 + r^2 (\mathrm d\theta^2 + \sin^2 \theta~\mathrm d\...

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I can answer some of it, and in such a way that it has invariant general relativistic meaning. However, not a general answer. You do have to, and can, treat curvature and some measures of volume invariantly. There are two questions. 1)Does negative/positive curvatures have more volume, that some (in some sense) equivalent spacetime with no curvature? And 2)...

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In a question like this you need to ask what does the volume change relative to. So it's a little bit ambiguous. However, the answer to your question is "yes" in the following restricted sense. Imagine having a "swarm" of test objects, with mass so small that their effect on the spacetime around them is negligible. Assume that they are in freefall, i.e. ...

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