Recent content by tag16

Basically, I'm asking if you were given a PV graph similar to this: http://www.websters-online-dictionary.org/images/wiki/wikipedia/commons/thumb/d/dc/Stirling_Cycle.png/200px-Stirling_Cycle.png How would you find ΔG and ΔA for the processes 1 to 2, 2 to 3, 3 to 4 and 4 to 1?

For a thermodynamic process, what equations would be used to find the change in Gibbs and Helmholtz free energy when: a.)The process is adiabatic b.)The process is isothermic c.)The process is at constant volume d.)The process is at constant pressure I know ΔG=ΔH-TΔS and ΔA=ΔU-TΔS but do...

I tried 1/2 when I was originally having difficulty with the problem and still got the wrong answer. I guess I must be missing something else.

I'm not sure, I thought maybe 1/8 but that's not right.

Homework Statement Evaluate the double integral by converting to polar coordinates. ∫∫ arctan y/x dA; R is the sector in the first quadrant between the circles 1/4= x^2+y^2 and x^2+y^2=1 and the lines y=x/√3 and y=x. Homework Equations arctan y/x= θ The Attempt at a Solution...

Is this right? If not I don't know how to find it. PV = nRT --> n =PV/RT--> n = (1 − α)n + 2αn = PV/RT α = (PV/nRT)− 1

x + y = 1 92.011 x + 46.0055 y = 77.4395 x= -y + 1 92.011 (-y +1) + 46.0055 y = 77.4395 -92.011 y + 92.011 + 46.0055 y = 77.4395 -46.0055 y + 92.011= 77.4395 -46.0055 y = -14.5715 y = 0.316734 1- y = x x = 0.683266 To find the degree of dissociation I did this...

opps I think I used the wrong gas constant again: M=(0.749g/254.882 cm^3)(82.057)(303.15 K)/(0.94396 atm) ave. molecular mass of bulb 1: 77.4395 g

For bulb 1: P=717.41 Torr= 0.94396 atm V= 254.882 cm^3 T=303.15 K mass= 141.9780 g(bulb w/gas)-141.229 g(bulb empty)= 0.749 g M=(0.749g/254.882 cm^3)(8.314)(303.15 K)/(0.94396 atm) ave. molecular mass of bulb 1: 7.846g

so I used M=DRT/P to find the average molecular mass and got this: For bulb 1: P=717.41 Torr= 0.94396 atm V= 254.882 cm^3 T=303.15 K mass= 141.9780 g M=(141.9780 g/254.882 cm^3)(0.0821)(303.15 K)/(0.94396 atm) ave. molecular mass of bulb 1: 14.686 Am I going about this the right...

I'm not really sure what to do next, I think I'm suppose to use this equation: M(mix)=x(NO2)M(NO2)+x(N2O4)M(N2O4), if so I'm not sure how to get the mole fractions from the data I have or the degree of dissociation.

opps that's what I did originally but I also averaged them all together which was wrong. This is minus the mass of the bulb: bulb1:254.882 bulb2:263.277 bulb3:284.049

So the volume of the bulbs using V= mass (w/water)/ water density bulb 1: 397.122 bulb 2: 362.841 bulb 3: 398.928 Which brings me back to my problem of finding the average molecular weight of the gas mixture at each temperature. I originally thought I was suppose to use this equation...

I just assumed what I originally thought the volume was must be wrong or you wouldn't asked about the volume so I just came up with something random because I didn't know what else it could be, hence the 250 cm^3. So I have to do the calculations for each of the three bulbs separately as well...

In a previous post I wrote this: Since in a later post you asked me how to get the volume, I just assumed that what I had already posted must have been wrong. When I calculated it I got V=267.403 cm^3. To get the mass to use in this equation would I want to take the average of the three...