Find the ratio (ideal gas laws) how to get a Ratio between two numberS>

[teggenspiller]

Homework Statement

The temperature of a fixed mass of gas drops from 127 C to 27 oC under constant pressure. Find the ratio of the new volume to the original volume.

C. 3:4
D. 4:3
E. 1:1
A and B arent here because i have gotten them wrong.

Homework Equations

Ideal gas law: PV= nRT

Pressure in Pa, Volume in Liters, n as in moles, R is a constant (0.0821) T= Temp Kelvin
Looking for the ratio of the final V to the initial V

The Attempt at a Solution

P= constant, n= constant, R= constant, so V= nRT/P and if p, n and R are constant then V just = T then right?

so V/T= V/T

And V/ 127 = V/27


what do i do here? 127/27 = 4.7

how do i get a 'ratio'?? between the two?

 

[fss]

What if the question read,

The temperature of a fixed mass of gas drops from 100 C to 0 C under constant pressure.

Would your method of dividing T1 by T2 work in this case using Celsius?

 

[gneill]

Homework Statement

The temperature of a fixed mass of gas drops from 127 C to 27 oC under constant pressure. Find the ratio of the new volume to the original volume.

C. 3:4
D. 4:3
E. 1:1
A and B arent here because i have gotten them wrong.

Homework Equations

Ideal gas law: PV= nRT

Pressure in Pa, Volume in Liters, n as in moles, R is a constant (0.0821) T= Temp Kelvin
Looking for the ratio of the final V to the initial V

The Attempt at a Solution

P= constant, n= constant, R= constant, so V= nRT/P and if p, n and R are constant then V just = T then right?

so V/T= V/T

And V/ 127 = V/27what do i do here? 127/27 = 4.7

how do i get a 'ratio'?? between the two?

When you write your equations, clearly label the "before" and "after" variables. Numbering them is usual the way: P1, V1, T1, P2, V2, T2, and so on.

You are looking for a ratio of the new volume to the old volume. So you want to find V2/V1.

You've already written "V/T = V/T" which I must presume means

V1/T1 = V2/T2

So rearrange to find V2/V1. Be sure to use absolute temperatures!

 

[teggenspiller]

no i don't think it would.. (?)

 

[teggenspiller]

the dividing part wouldn't work. (physically)

but the ratio would be 0:100 or 100:0

?

 

[teggenspiller]

wow. gniel. thanks so much! KELVING temps. duurrrh

 

[teggenspiller]

so it would be 4:3. since the volume of first temp would be less than of second? (since the Temp Values and they have to even out?)

 

[gneill]

so it would be 4:3. since the volume of first temp would be less than of second? (since the Temp Values and they have to even out?)

It would be good to get an intuitive feel for the result. You have a certain quantity of gas at a given temperature and pressure. Now you cool it down. What happens as gas cools? The volume shrinks and/or the pressure drops. To keep the pressure constant as the problem states, the the volume must be made smaller. So you must expect that the ratio of new volume to old volume will be less than one: V2/V1 < 1.

 

[teggenspiller]

It would be good to get an intuitive feel for the result. You have a certain quantity of gas at a given temperature and pressure. Now you cool it down. What happens as gas cools? The volume shrinks and/or the pressure drops. To keep the pressure constant as the problem states, the the volume must be made smaller. So you must expect that the ratio of new volume to old volume will be less than one: V2/V1 < 1.

see that's where my concepts are all messed up.. Because when I try to get a feel for it i think of V/T and I think that if temp gets smaller, the volume has to increase to keep the ratios the same.

but now that you mention, if you cool a balloon by 100K, the Volume will DEFINITLY decrease to keep the pressure the same.

 

[teggenspiller]

I keep examining what u wrote about temp going down means volume and/or pressure goes down. What if pressure is Dropped and Volume is INcreased?? then the temp will stay the same if they are decreased and increased by an even amount?

Ex: if Pressure is dropped by say, 1/3rd and Volume is Doubled or Quadrupled, Then the temp would go down with the drop in pressure, but back UP with the V increase??

 

[gneill]

I keep examining what u wrote about temp going down means volume and/or pressure goes down. What if pressure is Dropped and Volume is INcreased?? then the temp will stay the same if they are decreased and increased by an even amount?

Ex: if Pressure is dropped by say, 1/3rd and Volume is Doubled or Quadrupled, Then the temp would go down with the drop in pressure, but back UP with the V increase??

P₁*V₁ = nR*T₁ and P₂*V₂ = nR*T₂

Put them together in ratios

$$\frac{P_2 V_2}{P_1 V_1} = \left(\frac{P_2}{P_1}\right) \cdot \left(\frac{V_2}{V_1}\right) = \frac{T_2}{T_1}$$

You should be able to tell what happens to the ratio T2/T1 when you change the ratios of pressures and volumes.