Physics/Chemistry guru's (or novice's)

[Vegitto]

Hey, all .

I've got a chemistry test in 3 hours, and I was cramming. It's 6:06 over here, I got up at 5:15. I've studied a lot already, but I really need this grade to make up for other grades. I'm not a genius .

I've got one question on this chapter, which is about moles, gas pressure and the likes. I know that according to Gay-Lussac's (and Charles') Law, you can calculate volume or temperature, given that you know one whole fracture and one partial, ie, 1 ml = 273K, 2 ml = xxx K, by using either V1/T1 = V2/T2, T1/V1 = T2/V2 or V1T2 = V2T1.) What I need to know now, is the temperature of a gas with a certain pressure.

For example:
Ammonia's standard density is 0.77 kg/m3, it's temperature is 273 K, the pressure is p0. I've got a new 'tube' of ammonia, but this time, it's density is 0.68 kg/m3 and the pressure is still p0.

I couldn't find how to calculate the temperature of this gas, and I hope you can explain it to me .

I Googled a lot, and I know that Charles' Law, Gay-Lussac's Law, Avogadro's Law, Boyle's Law and the Ideal Gas Law don't have the answer.

Thanks in advance .

 

[hypn0tik]

Combined gas law perhaps?

 

[Howard]

I used to use pV = nRT a lot.

 

[Vegitto]

Originally posted by: hypn0tik
Combined gas law perhaps?

Pressure and volume are equal in both cases.

 

[hypn0tik]

Originally posted by: Vegitto

Originally posted by: hypn0tik
Combined gas law perhaps?

Pressure and volume are equal in both cases.

Well, if it's a closed system, the volume cannot be the same due to the conservation of matter. If nothing is able to flow in or out, you simply can't 'lose' matter. Hence, the volume cannot be a constant. It will increase to offset the lower density.

If that's not correct then I don't have a clue. Chemistry isn't really my forte.

 

[Vegitto]

Originally posted by: hypn0tik

Originally posted by: Vegitto

Originally posted by: hypn0tik
Combined gas law perhaps?

Pressure and volume are equal in both cases.

Well, if it's a closed system, the volume cannot be the same due to the conservation of matter. If nothing is able to flow in or out, you simply can't 'lose' matter. Hence, the volume cannot be a constant. It will increase to offset the lower density.

If that's not correct then I don't have a clue. Chemistry isn't really my forte.

I have one seperate 'box' that contains 1 m3 of a gas, at a pressure of p0 and a density of 0.77 kg/m3. In other words, the contents of the 'box' are 0.77 kg. The temperature is 273 K
In another seperate 'box', I have another 1 m3 of gas, at an equal pressure of p0, but the density is 0.68 kg/m3, the contents of the 'box' are 0.68 kg. What is the temperature?

(I'm sure it's possible, it's just a gas with a lower density than usual.)

 

[Indolent]

309...

assume you have one m^3 of each gas

1/(m1*t1)=1/(m2*t2)

1/(.77*273)=1/(.68*T2)


seems like it works for me. I'm just thinking quickly though, it's been a while


It's just the combined gas law using mass instead of moles, pressure, volume, and r cancel.

 

[Vegitto]

Hmm, I've got two formulas, but both render different results..

Both ideal gas laws:

n(R)T = n(R)T (R doesn't have to be used, it's a constant) yields a temperature of 309 K.
N(k)T = N(k)T (k doesn't have to be used, it's a constant) yields a temperature of 435 K.

n = amount of gas in moles.
T = absolute temperature.
N = number of particles.
(R) = ideal gas constant.
(k) = Boltzmann constant.

Which one is correct, and what went wrong where?

 

[hypn0tik]

Originally posted by: Vegitto

Originally posted by: hypn0tik

Originally posted by: Vegitto

Originally posted by: hypn0tik
Combined gas law perhaps?

Pressure and volume are equal in both cases.

Well, if it's a closed system, the volume cannot be the same due to the conservation of matter. If nothing is able to flow in or out, you simply can't 'lose' matter. Hence, the volume cannot be a constant. It will increase to offset the lower density.

If that's not correct then I don't have a clue. Chemistry isn't really my forte.

I have one seperate 'box' that contains 1 m3 of a gas, at a pressure of p0 and a density of 0.77 kg/m3. In other words, the contents of the 'box' are 0.77 kg. The temperature is 273 K
In another seperate 'box', I have another 1 m3 of gas, at an equal pressure of p0, but the density is 0.68 kg/m3, the contents of the 'box' are 0.68 kg. What is the temperature?

(I'm sure it's possible, it's just a gas with a lower density than usual.)

I'm still not sure why you can't use the ideal gas law.

P1V1 = n1*R*T1
p0*V = (0.77*V/MM(NH3))*R*273

P2V2 = n2*R*T2
p0*V = (0.68*V/MM(NH3))*R*T2

The left side of both equations is the same:

0.77*V*R*273/MM(NH3) = 0.68*V*R*T2/MM(NH3)

=> 0.77*273 = 0.68*T2

Edit: Assuming Ammonia is NH3? Maybe NH4.

Also, beaten. I get 309K as well.

 

[Vegitto]

Originally posted by: Indolent
309...

assume you have one m^3 of each gas

1/(m1*t1)=1/(m2*t2)

1/(.77*273)=1/(.68*T2)


seems like it works for me. I'm just thinking quickly though, it's been a while


It's just the combined gas law using mass instead of moles, pressure, volume, and r cancel.

Using this, if it's correct, you can just use mass x temperature, too, right? 1/(mass x temperature) yields the same results, right?

 

[Indolent]

Originally posted by: Vegitto

Originally posted by: Indolent
309...

assume you have one m^3 of each gas

1/(m1*t1)=1/(m2*t2)

1/(.77*273)=1/(.68*T2)


seems like it works for me. I'm just thinking quickly though, it's been a while


It's just the combined gas law using mass instead of moles, pressure, volume, and r cancel.

Using this, if it's correct, you can just use mass x temperature, too, right? 1/(mass x temperature) yields the same results, right?

yeah it's the same thing, I just didn't simplify enough.

 

[Vegitto]

Originally posted by: hypn0tik

Originally posted by: Vegitto

Originally posted by: hypn0tik

Originally posted by: Vegitto

Originally posted by: hypn0tik
Combined gas law perhaps?

Pressure and volume are equal in both cases.

Well, if it's a closed system, the volume cannot be the same due to the conservation of matter. If nothing is able to flow in or out, you simply can't 'lose' matter. Hence, the volume cannot be a constant. It will increase to offset the lower density.

If that's not correct then I don't have a clue. Chemistry isn't really my forte.

I have one seperate 'box' that contains 1 m3 of a gas, at a pressure of p0 and a density of 0.77 kg/m3. In other words, the contents of the 'box' are 0.77 kg. The temperature is 273 K
In another seperate 'box', I have another 1 m3 of gas, at an equal pressure of p0, but the density is 0.68 kg/m3, the contents of the 'box' are 0.68 kg. What is the temperature?

(I'm sure it's possible, it's just a gas with a lower density than usual.)

I'm still not sure why you can't use the ideal gas law.

P1V1 = n1*R*T1
p0*V = (0.77*V/MM(NH3))*R*273

P2V2 = n2*R*T2
p0*V = (0.68*V/MM(NH3))*R*T2

The left side of both equations is the same:

0.77*V*R*273/MM(NH3) = 0.68*V*R*T2/MM(NH3)

=> 0.77*273 = 0.68*T2

Edit: Assuming Ammonia is NH3? Maybe NH4.

Also, beaten. I get 309K as well.

So, the n(R)T = n(R)T is correct? What went wrong with the other one?

Thanks a lot, guys .

 

[hypn0tik]

Originally posted by: Vegitto

Originally posted by: hypn0tik

Originally posted by: Vegitto

Originally posted by: hypn0tik

Originally posted by: Vegitto

Originally posted by: hypn0tik
Combined gas law perhaps?

Pressure and volume are equal in both cases.

Well, if it's a closed system, the volume cannot be the same due to the conservation of matter. If nothing is able to flow in or out, you simply can't 'lose' matter. Hence, the volume cannot be a constant. It will increase to offset the lower density.

If that's not correct then I don't have a clue. Chemistry isn't really my forte.

I have one seperate 'box' that contains 1 m3 of a gas, at a pressure of p0 and a density of 0.77 kg/m3. In other words, the contents of the 'box' are 0.77 kg. The temperature is 273 K
In another seperate 'box', I have another 1 m3 of gas, at an equal pressure of p0, but the density is 0.68 kg/m3, the contents of the 'box' are 0.68 kg. What is the temperature?

(I'm sure it's possible, it's just a gas with a lower density than usual.)

I'm still not sure why you can't use the ideal gas law.

P1V1 = n1*R*T1
p0*V = (0.77*V/MM(NH3))*R*273

P2V2 = n2*R*T2
p0*V = (0.68*V/MM(NH3))*R*T2

The left side of both equations is the same:

0.77*V*R*273/MM(NH3) = 0.68*V*R*T2/MM(NH3)

=> 0.77*273 = 0.68*T2

Edit: Assuming Ammonia is NH3? Maybe NH4.

Also, beaten. I get 309K as well.

So, the n(R)T = n(R)T is correct? What went wrong with the other one?

Thanks a lot, guys .

Nothing. You probably punched it into your calculator incorrectly. Look at what the PV = nRT simplified to. It essentially became N1T1 = N2T2.

 

[hiredgoons]

PV = nRT
PV/(nRT) = 1
P1V1/(n1R1T1) = P2V2/(n2R2T2)

P, V and R are the same in both cases

n1T1 = n2T2

Moles and mass are proportional, therefore you can substitute mass for n and the result will be the same:

m1T1 = m2T2

 

[Vegitto]

Originally posted by: hypn0tik

Originally posted by: Vegitto

Originally posted by: hypn0tik

Originally posted by: Vegitto

Originally posted by: hypn0tik

Originally posted by: Vegitto

Originally posted by: hypn0tik
Combined gas law perhaps?

Pressure and volume are equal in both cases.

Well, if it's a closed system, the volume cannot be the same due to the conservation of matter. If nothing is able to flow in or out, you simply can't 'lose' matter. Hence, the volume cannot be a constant. It will increase to offset the lower density.

If that's not correct then I don't have a clue. Chemistry isn't really my forte.

I have one seperate 'box' that contains 1 m3 of a gas, at a pressure of p0 and a density of 0.77 kg/m3. In other words, the contents of the 'box' are 0.77 kg. The temperature is 273 K
In another seperate 'box', I have another 1 m3 of gas, at an equal pressure of p0, but the density is 0.68 kg/m3, the contents of the 'box' are 0.68 kg. What is the temperature?

(I'm sure it's possible, it's just a gas with a lower density than usual.)

I'm still not sure why you can't use the ideal gas law.

P1V1 = n1*R*T1
p0*V = (0.77*V/MM(NH3))*R*273

P2V2 = n2*R*T2
p0*V = (0.68*V/MM(NH3))*R*T2

The left side of both equations is the same:

0.77*V*R*273/MM(NH3) = 0.68*V*R*T2/MM(NH3)

=> 0.77*273 = 0.68*T2

Edit: Assuming Ammonia is NH3? Maybe NH4.

Also, beaten. I get 309K as well.

So, the n(R)T = n(R)T is correct? What went wrong with the other one?

Thanks a lot, guys .

Nothing. You probably punched it into your calculator incorrectly. Look at what the PV = nRT simplified to. It essentially became N1T1 = N2T2.

Oh, wait, I didn't quite get Avogadro's number correct . I think I had something around .02212 x 10^23 instead of 6.02212 x 10^23 ..

Thanks a lot, guys!
I really feel like I'm gonna nail this test, now.

 

[hypn0tik]

Good luck!

 

[Vegitto]

Originally posted by: hypn0tik
Good luck!

Thanks, man . When I get back, I'll let you guys know how I did .

 

[Vegitto]

Hey, wait a minute..

Is there a law that states that for a gas, PV/MT = c?
Because this works with all gasses.. I couldn't find a law like that. I'm probably not the first to think of it .. But if I am, man, that rocks!

EDIT: My new law seems to be impractical, there is a different constant for each gas. For example, in ammonia (NH3) the constant is 475,714, but for carbon dioxide (CO2), the constant is 184,441..

Somehow, this 'law' is really handy and really unhandy at the same time .

 

[hypn0tik]

Originally posted by: Vegitto
Hey, wait a minute..

Is there a law that states that for a gas, PV/MT = c?
Because this works with all gasses.. I couldn't find a law like that. I'm probably not the first to think of it .. But if I am, man, that rocks!

Sure, if you re-arrange the ideal gas law to read: PV/nT = R, where R is your constant.

 

[Vegitto]

Originally posted by: hypn0tik

Originally posted by: Vegitto
Hey, wait a minute..

Is there a law that states that for a gas, PV/MT = c?
Because this works with all gasses.. I couldn't find a law like that. I'm probably not the first to think of it .. But if I am, man, that rocks!

Sure, if you re-arrange the ideal gas law to read: PV/nT = R, where R is your constant.

Oh.. Oops .. But this system (heh, I've got my own system now ) uses moles instead of kilograms..

On the other hand, using moles DOES make it a real constant, no matter what gas it is.. But this is just a tiny bit easier (depending on your variables, of course)..

I think we can conclude that I am an idiot, but at a slightly higher level than a normal idiot .

 

[Vegitto]

Yet another double post, but meh :

nRT = NkT

n = N / Avogadro's number
R = k x Avogadro's number
T = T

Why are these recognized as different laws when they're the same? That's like A = B and 2A = 2B, right?

 

[hiredgoons]

You can derive any of the gas laws from PV = nRT, but they're recognized as separate anyway. Most equations related to linear motion in Newtonian Physics are the same way. It probably has to do with when they were discovered.

 

[Vegitto]

Originally posted by: hiredgoons
You can derive any of the gas laws from PV = nRT, but they're recognized as separate anyway. Most equations related to linear motion in Newtonian Physics are the same way. It probably has to do with when they were discovered.

Ah, okay . Thanks.

On another note, p (momentum) = mv, and E = .5mv^2.

I think that p and E are congruent.. Anyone know if this is true, and if so, what the formula is?

Yeah, I've got a lot of questions.

 

[hiredgoons]

I'm not sure what you mean by congruent, but you could use E = pv/2. I'm not sure if this helps or not.

 

[Vegitto]

Originally posted by: hiredgoons
I'm not sure what you mean by congruent, but you could use E = pv/2. I'm not sure if this helps or not.

You mean E = (PV)/2? I'll try, thanks .

It's not really important for a test or anything, I was just curious. Who made this law, if you don't mind me asking?

EDIT: Excellent, it works! How does this work, mathematically? I'm really full of questions, today .

In other words, why is:

a = xy
b = .5xy^2
b = (ay)/2
a = (2b)/y

How did you (or they, whatever ) do that?