Recent content by Dazed&Confused

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    Specified equation of state from heat capacity

    Maybe it was too late for me to be posting before but my first method misses the extra constant you'd get.
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    Specified equation of state from heat capacity

    That's definitely a good idea! Integrating c_v/T we find s = \frac{AT^3}{3} +f(v) for some f. Then with the Maxwell relation f'(v) = \frac{B'(T)}{v-v_0} which means that B'(T) = E for some E. So this confirms what I found before (much more easily).
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    Specified equation of state from heat capacity

    Homework Statement The constant-volume heat capacity of a particular simple system is c_v = AT^3 where A is a constant. In addition the equation of state is known to be of the form (v-v_0)p = B(T) where B(T) is an unspecified function of T. Evaluate the permissible functional form of B(T)...
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    Why Does the Compressibility of Helium Gas Diverge in This System?

    So in my answer I found that the compressibility is infinite if \mu, T is fixed which has to be the case if \mu is determined by T and p only. A constant T and \mu] implies a constant p irrespective of V. I'm just not sure if that is the answer they want and what the physical significance of it...
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    Why Does the Compressibility of Helium Gas Diverge in This System?

    I feel silly now as I wrote d\mu =-sdT +vdp . If the system is equilibrium should not dU + dU_r = 0 = (T-T_r)dS + (\mu-\mu_r)dN mean that T=T_r and \mu_r = \mu if the system undergoes a quasi static change? I guess that would 'explain' why I found the derivative to be infinite as the...
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    Why Does the Compressibility of Helium Gas Diverge in This System?

    Thank you for the response. I agree with you on your first point but I do not understand why the chemical potential matches the reservoir only if the pressure is constant. I thought the condition for equilibrium would set the chemical potential to be the reservoir's regardless, and should not...
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    Why Does the Compressibility of Helium Gas Diverge in This System?

    Homework Statement A cylinder is fitted with a piston, and the cylinder contains helium gas. The sides of the cylinder are adiabatic, impermeable, and rigid, but the bottom of the cylinder is thermally conductive, permeable to helium, and rigid. Through this permeable wall the system is in...
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    Equilibrium volume of two differential van der Waal gases

    Thanks! I notice that if in the denominator the as and b are ignored, then my final equation matches yours.
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    Equilibrium volume of two differential van der Waal gases

    Ok this time I expanded around V/2 like you said :P and got V_1 = V/2 + \frac{4(a_2-a_1)}{\frac{16(a_2+a_1)}{V} +\frac{2RTV^2}{(V/2-b)^2}}
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    Equilibrium volume of two differential van der Waal gases

    Thanks for the response. First I should say I didn't mean to write the N as it is 1, although it doesn't really matter. I think I understand what you mean, but I am apprehensive about the right hand side expansion, unless you mean something like \frac{RT}{V_1} + \frac{RTb}{V_1^2} -...
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    Equilibrium volume of two differential van der Waal gases

    Homework Statement Two ideal van der Waals fluids are contained in a cylinder, separated by an internal moveable piston. There is one mole of each fluid, and the two fluids have the same values of the van der Waals constants b and c; the respective values of the van der Waals constant ''a'' are...
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    Van der Waal expansion and delivered work

    I see my problem: there are two Qs. The first equation was the heat from the subsystem and the energy and entropy the heat RHS.
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    Van der Waal expansion and delivered work

    Homework Statement Assume that one mole of an ideal van der Waals fluid is expanded isothermally, at temperature T_h from an initial volume V_i to a final volume V_f. A thermal reselvoir at temperature T_c is available. Apply dW_{RWS} = \left ( 1 - \frac{T_{RHS}}{T} \right ) (-dQ) +(-dW) to a...
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    Quantum Theory: derive EoM of action for a 'general' potential

    This is the beginning of writing it out as two derivatives, the physicist's way. Simply divide and multiply each by \delta \phi and \delta \phi^* respectively and take the infinitesimal limit.
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