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A. V1/V2 = T2/T1
B. V1/T1= V2/T2
C. V2= (T2/T1) V1
D. V1 T1= V2 T2
E. none of the above

Thanks in advance, I just don't get it.

2006-06-14 08:05:09 · 7 answers · asked by izzynindie 2 in Science & Mathematics Chemistry

thanks to those helping...but it's not that i don't understand the BASIC math, duh, I am just confused, and i didn't actually ask for opinions. Thanks.

2006-06-14 08:24:17 · update #1

7 answers

D. Based on the ideal gas law, PV=nRT

Solve for R to get: R = (P*V)/(n*T)

for two different conditions, then, you would have:

(P1*V1)/(n1*T1) = R = (P2*V2)/(n2*T2) or,

(P1*V1)/(n1*T1) = (P2*V2)/(n2*T2)

or P1*V1*n2*T2 = P2*V2*n1*T1

Note how V1 and T1 can never be in the numerator together?

In order for the ratio to work, one must be the numerator, the other the denominator, because volume is directly proportional to temperature -- the higher the temperature (keeping number of moles and pressure equal), the larger the volume.

2006-06-14 09:25:02 · answer #1 · answered by Dave_Stark 7 · 2 0

During the first two weeks of any thermodynamic class, students are taught the "ideal gas law" which dictates how an idealized gas will behavior, in terms of temperature, pressure, and volume. The remaining semester teaches that real-world gases seldom behave as such. It sometimes comes close, but some (esp. steam never behave ideally). It's components were experimentally determined by Jacques Charles in 1787, and Robert Boyle in 1662. Charles was an especially cool dude, in spite of being French :-)

Assuming that the mass of gas remains the same (that is, it is a closed system),

PV/T (at the first state) = PV/T (at the second state)

Thus, if you increase the temperature, the pressure (or volume) must also increase to keep the value (PV/T) the same. In your equations above they are ignoring pressure, which is weird, so I assume they are keeping it constant.

so,

a. v1/t1 = v2/t2 ...... rearangement.... v1/v2 = t1/t2... this is bad
b. v1/t1 = v2/t2 ...... rearangement.... v1/t1 = v2/t2 .. this is ok
c. ....................................................... v2 = (t2/t1)*v1.. this is ok
d. ....................................................... v1t2 = v2t1 ... this is bad

So this is strange, looks like both a, and d, are not accurate -- rearanging the terms p1v1/v1=p2v2/t2 does not match what you have listed. Are you sure about those multiple choice questions?

By the way, (a) is definately wrong.

v1/t1 = v2/t2

v1/(t1*v2) = (1/t2)

(v1/v2) = (t1/t2) which does not match (a). Think of it this way, volume1 goes up, the temperature also needs to increase to keep the equality. (a) says that if volume increases, the temp decreases, which is just flat out wrong. As another example, if V2 is twice that of V1, that implies that the temperature must also be much higher, not the other way around.

For subsequent postings, someone show me how (a) is correct. You can't of course. Argh.... can't people do algebra here!!!!! Trust me, as the answer is written both (a) and (d) are wrong.

2006-06-14 08:36:15 · answer #2 · answered by Boffin 2 · 0 0

PV=nRT
P=pressure
V=Volume
n=number of moles
R=gas constant for ideal gases
T= Temperature
(P1V1)/(P2V2)=(n1RT1)/(n2RT2)
As long as pressure and number of moles are constant this means P1=P2 and n1=n2: and they therefore drop out of the equation along with R. Which gives:

V1/V2=T1/T2
If you multiply both sides by V2/T1, you get:

V1/T1=V2/T2

If you multiply both sides of this equation by T2 you get:

T2(V1/T1)=V2 which re-written is

V2=(T2/T1) V1

The only equation you can't get is D.
V1 T1= V2 T2

2006-06-14 08:43:16 · answer #3 · answered by the_mad_scientist 1 · 0 0

Start with the Ideal Gas law in two separate states, where pressure and number of moles is the same.

PV1 = nRT1
PV2 = nRT2

Divide Eq. 1 by Eq. 2 and then manipulate to get the possible solutions. D is the only one you can't do, unless the conditions at State 1 and State 2 are the same.

2006-06-14 08:19:51 · answer #4 · answered by TheOnlyBeldin 7 · 0 0

start playing with the equation.

(P1V1)/T1=(P2V2)/T2

it all comes down to basic algebra. i hate to say this, but if you're problem is with the math you need to review this before you go any further or you're gonna end up in real trouble...

2006-06-14 08:20:52 · answer #5 · answered by shiara_blade 6 · 0 0

E - 99 cents per gallon is now impossible for ideal gas.

2006-06-14 08:08:43 · answer #6 · answered by Anonymous · 0 0

Look at this website.
http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/idegas.html
These are just algebraic rearraingements of the variables. See which one is not the same equation.

LoAnnie81

2006-06-14 08:14:46 · answer #7 · answered by LoAnnie81 3 · 0 0

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