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Let $m_L$ represent the mass of liquid in the tank at equilibrium, $m_V$ represent the mass of vapor in the tank at equilibrium, $v_L$ represent the specific volume of the liquid, $m_V$ represent the volume … of vapor, and V represent the tank volume. …

If the mass m is constant and the tank volume V is constant, then $v_{avg}=\frac{V}{m}$ is constant for any changes that occur within the tank. …

The specific volume they have calculated is the average value for the contents of the vessel, weighted in terms of mass fraction of liquid and mass fraction of vapor. …

Suppose that, at time t, we were to supplement the material in the control volume by the tiny amount of mass that is about to enter between times t and $t+\Delta t$. … against the "piston" formed by the material ahead of it minus the work to push material into the control volume by the "piston" formed by the material behind it. …

The specific volume of a saturated liquid also decreases with pressure (at constant temperature) as you move into the sub cooled liquid zone. …

For an ideal solution, the molar volume of the solution is a weighted average (in terms of mole fractions) of the molar volumes of the pure components at the same temperature and pressure as the mixture … : $$V^{ID}=V^0_Ax_A+V^0_Bx_B=V^0_A+(V^0_B-V^0_A)x_B$$The excess molar volume is equal to the actual molar volume minus the ideal molar volume: $$V^{EX}=V_m-V^{ID}$$or$$V_m=V^{ID}+V^{EX}$$ …

This is how to get the equation for the tangent line. The starting equations are $$[V_m]_{x=x_0}=(1-x_0)[V_A]_{x=x_0}+x_0[V_B]_{x=x_0}$$and$$\left[\frac{dV_m}{dx}\right]_{x=x_0}=[V_B]_{x=x_0}-[V_A]_{ …

For an ideal gas, the temperature would not change, and the volume increase would be exactly offset by a pressure decrease, such that Pv = constant. …

The article is trying to provide a physical interpretation to the enthalpy. All this ever does is confuse the student. Enthalpy is not a fundamental quantity like internal energy or entropy. It is …

$PV^{\gamma}=C$ applies only to the reversible expansion of an ideal gas. Free expansion is not a reversible expansion because, even though the gas can be returned to its original state, its surround …

v}$ and a viscous normal stress component $\tau$, where T, v, and P vary with position dA at the piston, and the viscous stress component varies not with the amount of gas deformation (local specific volume …

In your example, the isotropic part $\sigma _I$ is $$\sigma_I=\frac{(\sigma_z+\sigma_r+\sigma_{\theta})}{3}$$ and is responsible for volume changes. … The deviatoric part of the stress state is given by $\sigma_z-\sigma_I$, $\sigma_r-\sigma_I$, and $\sigma_{\theta}-\sigma_I$; the deviatoric part of the stress tensor is responsible for a volume-preserving …

To elaborate on the answer I gave, the state of stress for the deformed rod is $$\sigma_z=E\epsilon_z$$$$\sigma_r=0$$$$\sigma_{\theta}=0$$and the strains are $$\epsilon_z=\frac{\sigma_z}{E}$$ $$\epsil …

$$dH=dU+pdV+Vdp=C_pdT+V(1-\alpha T)dP$$So, $$dU=C_pdT-pdV-V\alpha Tdp$$The rest is basically what you've already done with your second method.

If we neglect the pressure difference across the balloon membrane, this work is equal to the outside air pressure times the change in volume of the balloon (assuming that the amount of air that enters … the balloon is not sufficient to deplete the volume of the balloon when the pressure in the chamber matches the outside air pressure). …