Recent content by ebunny91

Homework Statement A certain substance exists in two phases at equilibrium at Temp T1 and pressure P1. One phase α is a crystalline solid (density 0.531 g/cm^3) while the other phase β is an amorphous glass (density 0.510 g/cm^3). a) If the external pressure is increased, will the temp...

Yea I think I have to prove the Gibbs Phase Rule (somehow using something about ΔHfusion)

Homework Statement Prove that a on-component system must have a triple point. You may assume that ΔHfusion>0, if needed. Homework Equations C (components) = #of distinct substances - # of distinct chemical reactions Gibbs Phase Rule: degrees of freedom= components - phases + 2 or...

Homework Statement For both H20 and D20: Cp = 75.5 J/mole-k for the liquid Cp = 37.8 J/mole-k for the solid ΔHfus=6.01 kJ/mole. Liquid H20 and D20 form an ideal solution. a) One mole of liquid H20 at 10 C is mixed with four moles of LIQUID D20 at 0 C in an insulated...

Homework Statement When two moles of CO are introduced into a vessel containing solid sulfur, the final equilibrium pressure is 1.03 atm. Determine Kp for the reaction: S(s) + 2CO(g) <--> SO2(g) + 2C(s) Homework Equations Kp=P(products)/P(reactants) Mole Fraction = mole of A/...

Oh I forgot to convert the atm to joules/liter. Thanks!

So in equation 1, it's at T1 = 77K. Then by equating the equations I only have one unknown which is T2 and that turns out to be 77.00736044 K

I have two equations for the slope of a line: 351.5/T2 (1.47 - 1.23) and 0.59-0.45 atm/T2 - 77K Since it is the same line, the two equations (slopes) should be equal. So I equated them to solve for final temperature. The question tells me to assume that the final temperature is close to 77K...

But then I would only have the slope of the line. I have to determine the temperature at which the system must be brought to insure that the three phases are present. Do I have to do something more to get the temperature of the triple point?

Well maybe I can use the two equations to get the ΔH (a-->b) First eqn: Eb(a) ---> Eb(c) ΔH (a --> c) = 375.8 J/mole and flip/negate the second eqn: Eb(c) ---> Eb(b) ΔH (b --> c) = - 24.3 J/mole So Eb(a) ---> Eb(b) ΔH = ΔH (a --> c) + ΔH (b --> c)...

Oh ok, so dP/dT= ΔP/ΔT = P2-P1/T2-T1 = 0.59-0.45 atm/T2 - 77K and then maybe I can use dP/dT= ΔH/TΔV to get the slope of the same line? But I don't have ΔH (a-->b)...

It was a typo yea. It's supposed to be 77 K. I know that at the triple point all three solid phases should be in equilibrium at the same temperature and pressure. The problem already gives us the pressure (0.59 atm) so now we need to find the temperature at the triple point. (I'm just typing...

Homework Statement The metal Eborium (Eb) has three solid phases: a, b, c. At a pressure of 0.45 atm, the a and b phase coexist at 70 K (temperature). The molar volume of Eb(a) is 1.23 liter/mole and that of Eb(b) is 1.47 liter/mole. The heats of transformation are as follows: Eb(a) --->...