Calculating volume from adiabatic operation

[mister_tom]

trying to work through the exercise sheets that were given to us to prepare for exams when term starts up but I've come to a complete stop on this question and the following questions all seem to be based around it so i need your help!

Homework Statement

a certain gas occupies a bolume of 4m³ at a pressure of 110KN/m² absolute and a temperature of 20 deg celsius. if the gas is compressed adiabatically until the pressure is 770KN/m² absolute, find the new volume

the specific heat capacities for this gas are Cp = 995J/kg celsisus Cv = 705J/kg celsius

Homework Equations

N/A

The Attempt at a Solution

no equations were given but here's where I've got to so far:

changing units into SI

P1 = 110,000 pa
P2 = 770,000 pa
T1 = 293 K
Q = 0
\gamma = Cp/Cv = 1.41

the only equation i can find in my notes that i think could be used is PV\gamma=constant but i cannot work out how to calculate the final volume from this. any help would be appreciated.

thanks tom

 

[Redbelly98]

the only equation i can find in my notes that i think could be used is PV\gamma=constant but i cannot work out how to calculate the final volume from this. any help would be appreciated.

thanks tom

Welcome to Physics Forums Tom.

Yes, that's the equation to use here. If you know both P and V at some time, you can calculate what "constant" is.

 

[mister_tom]

Welcome to Physics Forums Tom.

Yes, that's the equation to use here. If you know both P and V at some time, you can calculate what "constant" is.

hi thanks for the welcome.

been thinking about this and with the equation posted if i knew both P and V at a given point then there would be no need to use the equation as id already have the volume at the given point. unless of course the constant at 1 point of volume was the same at another then the equation could be transposed to give volume with a given pressure and constant but i think I am barking up the wrong tree here.

have tried calculating both final volume and final temperature but without either I am stuck. I've looked at all the gas laws and i can't see how any can be used as none can be calculated with the info the question gives, the same goes for the ideal gas law. I've also looked at possibly transposing an equation with Q in as i know that to = 0 but again I am just hitting a brick wall.

is it possible for you to explain slightly further without giving anything away. I am at a compelete loss now

 

[Redbelly98]

Okay.

We know the pressure is 110 kPa when the volume is 4 m³. You can use that to calculate the constant,

P V ^γ​

Another useful way to set this up is to say

P₁ V₁^γ = P₂ V₂^γ
​

You can solve that equation for V₂, and then just plug in the numbers you know for P1, P2, V1, and γ

 

[mister_tom]

Okay.

We know the pressure is 110 kPa when the volume is 4 m³. You can use that to calculate the constant,

P V ^γ​

Another useful way to set this up is to say

P₁ V₁^γ = P₂ V₂^γ
​

You can solve that equation for V₂, and then just plug in the numbers you know for P1, P2, V1, and γ

ahh brilliant I am with you now, thanks for the help.

for some reason i didnt think it would be possible to calculate due to the temperature changing and trying to calculate it from that. looking back i actually wrote that equation down about 3 hours ago and crossed it out before i tied myself into a knot

the only thing I am not 100% on now is that the equation you just posted is referring to boyles which states that pressure must be constant, but with the adiabatic process temperature changes. is the reason that temperature dosent have to be taken into account is because the specific heat ratio is?

 

[Redbelly98]

the only thing I am not 100% on now is that the equation you just posted is referring to boyles which states that pressure must be constant, but with the adiabatic process temperature changes. is the reason that temperature dosent have to be taken into account is because the specific heat ratio is?

Nope, it's not Boyle's Law. That law says:

P₁ V₁ = P₂ V₂ ( at constant temperature )​

This equation is different, due to the γ superscript.