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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 …

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 …

At constant volume, $$dU=dH-VdP$$ …

If you know the initial and final states of the gas, it is easier to get the enthalpy change for a VDW gas by working with the internal energy change $\Delta U$ than $\Delta H$. This is because the i …

Your equation is wrong. It should read $$dH=\left(V-T\left(\frac{\partial V}{\partial T}\right)_P\right)dP$$ You may have to integrate this numerically because of the non-linearity with of the VDW eq …

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 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$$V_m=V^{ID}+V^{EX}$$ …

open-system (flow) version of the 1st Law of Thermodynamics, the work done by the system is subdivided into two parts: 1. the shaft work W and 2. the work to force fluid into and out of the control volume … In fact, wishin the control volume P and v are often not even spatially uniform. …

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 …

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. …

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. …

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. …

So, for an ideal gas, it doesn't matter whether the volume is varying or not, and we can always write for an ideal gas in that $$dU=nC_vdT\tag{ideal gas}$$So, for an adiabatic process of an ideal gas, …

If the process is adiabatic and reversible, then $$dS=\left(\frac{\partial S}{\partial T}\right)_VdT+\left(\frac{\partial S}{\partial V}\right)_TdV=0\tag{1}$$with $$\left(\frac{\partial S}{\partial T} …

In order to get the gas volume to increase at constant pressure, you need to add heat, which raises the temperature (average kinetic energy) of the gas. …