How to Calculate Temperature Change in Ideal Gas Law Problem for One Mole?

AI Thread Summary
The discussion focuses on calculating the temperature change of one mole of an ideal gas when its volume is doubled, using the equation P = P0 / {1 + (V0/V)^2}. Participants explore how to derive the temperature change in terms of constants P0, V0, and R, ultimately arriving at the answer of (11/10) P0V0/R. There is confusion regarding the interpretation of P0 and V0 as constants rather than initial conditions. The conversation emphasizes the importance of correctly identifying initial and final pressures and volumes to apply the ideal gas law accurately. The conclusion confirms that using V0 as the initial volume aligns with the provided solution.
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Homework Statement


for one mole of an ideal gas this relation holds good
P= P0 /{1+(V0/V)^2}, where P0and V0 are constants,
what will be its change in temperature in terms of P0,V0 and R,
if volume is doubled?


Homework Equations


for one mole of ideal gas we know PV=RT

Then RT/V= P0 /{1+(V0/V)^2}

How to proceed then? Can anyone help, please.


The Attempt at a Solution

 
Last edited:
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Evaluate the temperature of the gas when it occupies a volume V and 2V using the ideal gas law.
 
CAF123 said:
Evaluate the temperature of the gas when it occupies a volume V and 2V using the ideal gas law.

The Answer given in the text is(11/10) P0V0/R. How shall I reach?
 
Last edited:
According to the equation, if the initial volume is Vi, what is the initial pressure Pi? What is the initial temperature (using the ideal gas law? In terms of Vi, what is the final volume, Vf? What is the final pressure?

Chet
 
Chestermiller said:
According to the equation, if the initial volume is Vi, what is the initial pressure Pi? What is the initial temperature (using the ideal gas law? In terms of Vi, what is the final volume, Vf? What is the final pressure?

Chet

If we take initial pressure P0
and initial volume as V0
Initial temp = T1
Final Pressure = P0 (remaining constant)
final doubled volume =2V0
and final temp = T2

then by ideal gas law for one mole of gas
T2-T1=2P0V0/R -P0V0/R=P0V0/R

Is this a correct solution? Then what is the utility of the given equation?
 
dk_ch said:
If we take initial pressure P0
and initial volume as V0
Initial temp = T1
Final Pressure = P0 (remaining constant)
final doubled volume =2V0
and final temp = T2

then by ideal gas law for one mole of gas
T2-T1=2P0V0/R -P0V0/R=P0V0/R

Is this a correct solution?
No. If V0 is the initial volume, then, from the equation they gave, P0 is not the initial pressure. As they said in the problem statement, P0 and V0 are just constants for the problem (and are not related to the initial conditions).

Chet
 
for initial vol V0 the initial pressure is P0/2
for final vol 2V0 the final pressure is (4/5)P0
Then temperature change becomes =(11/10)P0V0/R

I think it is correct
 
dk_ch said:
for initial vol V0 the initial pressure is P0/2
for final vol 2V0 the final pressure is (4/5)P0
Then temperature change becomes =(11/10)P0V0/R

I think it is correct
Yes, if the initial volume is V0.

Chet
 
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Is it possible to arrive at the same result without having taken the initial volume as V0?
If possible please suggest a way out.
 
  • #10
It might be. The problem statement sort of implies this. Have you tried, and, if so, how far have you gotten?

Chet
 
  • #11
Chestermiller said:
It might be. The problem statement sort of implies this. Have you tried, and, if so, how far have you gotten?

Chet

I tried but failed to eliminate v1 initial volume to have the required relation. Please give hints if possible.
 
  • #12
I'll give it a shot.

Chet
 
  • #13
The only initial volume that matches the answer in the text is it is equal to Vo.

Chet
 
  • #14
So u have reached the same conclusion as I have , Thanks.
 

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