I will first enumerate the four points that need to be addressed.
The vapor and liquid phases have the same composition in an azeotrope.
The azeotrope boils at a constant temperature.
The composition of the azeotrope remains fixed while boiling.
The azeotropic mixture cannot be separated by fra...
Basically, no I can't explain the answer. Not without digging out my thermodynamics books and relearning all the stuff I've forgotten.
Ah, hang on, that's not the complicated answer we discussed yesterday. Let me have a read through it.
Hmm. OK. His first point is that we draw the vapour pressure plot with composition on the horizontal axis, and we normally put composition of the liquid on the horizontal axis.
But we could just as easily draw the same plot with composition of the vapour on the horizontal axis. Then the curve will have a different shape because in general the composition of the liquid and the vapour are different.
But I have to confess I'm not sure what he means by:
> This new vapor pressure curve must touch the original curve at the azeotropic composition x∗=y∗=z, for otherwise a tie line would connect two vapor phases at some P, which is physically unreasonable.
> This new vapor pressure curve must touch the original curve at the azeotropic composition x∗=y∗=z, for otherwise a tie line would connect two vapor phases at some P, which is physically unreasonable.
What is saying is plot both curves on the same diagram and they touch at three places (1) pure A, (2) pure B and (3) at the azeotropic composition.
But this is the bit I'm unsure about.
@Abcd a tie line is a horizontal line on the vapour pressure diagram i.e. a line of constant pressure. Specifically it's a horizontal line that joins the two curves $P(x)$ and $P(y)$.
I think it would be interesting to draw the two curves for an ideal mixture. We can do that because we can calculate $y$ from $x$ for an ideal solution. Give me a moment to think about it.
Which of the following must be true for adiabatic processes?
$C_\mathrm{v} = C_\mathrm{p}$
$\Delta H = 0$
$\Delta U = 0$
$\Delta S = 0$
$q = 0$
(Source: Chemistry GRE)
The answer is $q = 0$. From what I can find, an adiabatic process is when there is no transfer of heat,...
@Nobodyrecognizeable Projectile launched from a cliff : The equation for the trajectory is $y=xT-\frac{g(1+T^2)}{2v^2}x^2$ where $T=\tan\alpha$. Substitute $x=R, y=-H$. This enables you to eliminate $v^2$, which is the only unknown in the trajectory equation. Then as you suggested you can find maximum height by setting $\frac{dy}{dx}=0$.
@Nobodyrecognizeable The equation is a standard result. But you can derive it as John Rennie suggested, using the kinematic equations $x=ut$ and $y=vt-\frac12 gt^2$ where $u=V\cos\alpha$ and $v=V\sin\alpha$ and $V$ is launch velocity. Then eliminate $t$.
@sammygerbil I would have to integrate the velocity wrt time then I will get the displacement. I know if I make the y component of the displacement zero. Then I will get the time to fall and that times $v_0 cos \alpha t$ but is there some way if i wanna relate velocity and displacement ie mvdv/dx = -mg
@sammygerbil i am sorry as im typing in mobile. Im little late while replying.
@Nobodyrecognizeable All projectiles follow a parabolic path whatever their mass. So you do not need to consider the forces on the projectile. The horizontal and vertical motions are separate : the horizontal motion is constant velocity, the vertical motion is constant acceleration.
@Nobodyrecognizeable Sort of... My mother had dementia so I abandoned my research to look after her full time. But I was not doing very well anyway. I was not enjoying the research as much as I expected I would.
@Nobodyrecognizeable I don't have a job. I was a mature student so I was 43 when I started the PhD (2004). My mother died in February 2016. Now I do some volunteer work mending bicycles and I also speak to people about making life better in Scotland for people with dementia. I do not want to resume the research to finish the PhD.
Almost 100% are accepted, but not straight away. What happens is that you have an interview and you are asked questions about your research by 3 independent professors - usually one is an expert in the field.
At the interview they hand you a list of corrections for your thesis, and they discuss the important points with you. After you've made the corrections you can submit your thesis again, and it is usually accepted then.
Very occasionally they tell you that your research isn't good enough to get a PhD. They will suggest further research you could do, which might take another 3-12 months.
How do I solve this problem? The period of the free oscillations of the system shown here if mass M1 is pulled down a little and force constant of the spring is k and masses of the fixed pulleys are negligible, is
@Akash.B However before I do that I am working on a problem about the emf induced in a quarter-circular ring which rotates around a magnetic dipole. It looks tricky.
Once you're done with the current problem, I have one too: Consider a ball falling from a height $h=0.4m$ on an inclined plane of angle $\alpha=30^\circ$. It bounces back without losing velocity at an angle of $\beta=60^\circ$. The aim is to calculate the distance $d$ from the place of the first collision with the plane to the place of the second collision. I will show you a diagram and my attempt shortly.
@harambe Sorry I shall have to think some more about the problem because it is not coming out right for me either. I will ping you when I've solved it.
But you are given that the velocity doesn't change right after the collision
I mean... It's mostly kinematics IMO
The weird thing is that I somehow get $0$ with my technique, but the correct answer is $1.6m$. When I try replacing a cotangent with a tangent in my final formula, I get $1.6$ too which I find particularly weird
The angle of launch is 60 degrees, and the angle between the vertical and the downward incline is 120 degrees, so the launch angle bisects the angle between the vertical and the plane.
And according to expertsmind.com/learning/… the equation for maximum range down an inclined plane is $R=u^2/g(1-\sin\alpha)$ ... Whoops, just seen where I have made a mistake.
@harambe If you know the trick about launch angle for maximum range and the formula for maximum range up or down an incline, it can be solved fairly quickly. The angle of the incline is just right for this method to apply.
Calculate the enthalpy changr in KJ for 1 kmol water as it is vaporized at a const pressure of 101.3kPa.The specific voulmes of liquid and vapour at these conditions are 1.04X10-3 m3/kmol and 1.675 m3/kmol respectively.1030kJ of heat is added to water for this change
@harambe Well it's not the standard 9th grade material, it's mostly local / county (region) olympiad level (so below national, but not standard 9th grade curriculum)
@harambe Thank you! I have competitional experience already, got two bronze medals at nationals (both times less than 2 points below silver) but I am pushing a bit harder in the 9th grade because I like mechanics a lot and that's mostly all besides optics for this grade :)
@harambe goodbye. I will leave a note about the other problem later.
@gateprep The increase in volume is $1.675-0.00104m^3=1.674m^3$. So the work done against the atmosphere is $0.1013MPa \times 1.674m^3=0.17MJ$. So the total heat input (=enthalpy) is $40.7+0.17=40.87MJ$.
@gateprep If the vapour is expanding into the atmosphere ("at constant pressure"), this is the same as expanding into a piston cylinder with the atmosphere on the other side of the piston.
I will first enumerate the four points that need to be addressed.
The vapor and liquid phases have the same composition in an azeotrope.
The azeotrope boils at a constant temperature.
The composition of the azeotrope remains fixed while boiling.
The azeotropic mixture cannot be separated by fra...
Which of the following must be true for adiabatic processes?
$C_\mathrm{v} = C_\mathrm{p}$
$\Delta H = 0$
$\Delta U = 0$
$\Delta S = 0$
$q = 0$
(Source: Chemistry GRE)
The answer is $q = 0$. From what I can find, an adiabatic process is when there is no transfer of heat,...
