Calculation of pressure & volume both isothermally & adiabatically

AI Thread Summary
The discussion focuses on calculating pressure and volume in isothermal and adiabatic processes using given values for volume, pressure, and specific heat capacities. The isothermal calculation uses Boyle's law to derive a new volume, resulting in v2 = 0.285 m³. For the adiabatic process, the participants explore the relationship between pressure and volume using the adiabatic equality law, initially calculating v2 with a specific heat ratio (gamma) of 1.38 and later adjusting it to 1.5. There is a debate on the best method to solve for x in the equation x^n = y, with suggestions to avoid logarithms for simplicity. The conversation emphasizes the importance of understanding thermodynamic principles alongside mathematical techniques.
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Homework Statement
0.8 m3 of an experimental gas, compressed from a pressure of, 250 kn/m^2 to pressure of 7000 kn/m^2 . For gas in question, Cp = 1010 & Cv = 732.

Calculate the final volume of each case when gas is compressed.
>>> isothermally
>>> adiabatically

Calculate the final volume when gas is adiabatically compressed & ϒ= 1.5


p1 = 250 or 250000 n/m2
p2 = 7000 or 7000000 n/m2

Cp = 1010
Cv = 732
Relevant Equations
p1v1 = p2v2 [Boyle equality law (isothermal process)]

P1 V1^ Y= P2 V2 ^Y [adiabatic Equality law]
Hi, here is my attempt... Feel free to correct me if I am doing it wrong.
v1 = .8 m3
p1 = 250 or 250000 n/m2
p2 = 7000 or 7000000 n/m2

Cp = 1010
Cv = 7321.
isothermally
p1v1 = p2v2 [Boyle equality law (isothermal process)]
250 000 * .8 = 7000000 * v2
7000000 * v2 = 250 000 * .8
v2 = 250 000 * .8/ 7000 000
v2 = 0285 m3 <<<ans

adiabatically
Y = Cp / Cv
Y = 1010 / 732
y = 1.38

P1 V1^Y= P2 V2^Y [adiabatic Equality law]
250 000 * .8^ 1.38 = 7000000 * v2 ^1.38
7000000 * v2 ^1.38 = 250 000 * .735
7000000 * v2 ^1.38 = 183740.2
v2 ^1.38 = 183740.2/ 7000000
v2 ^1.38 = .0262 <<< what now?

2. Adiabatic again


y = 1.5 [new value for gamma]

P1 V1^Y= P2 V2^Y [adiabatic Equality law]
250 000 * .8^ 1.5 = 7000000 * v2 ^1.5
7000000 * v2 ^1.5 = 250 000 * .716
7000000 * v2 ^1.38 = 178885.43
v2 ^1.38 = 178885.43/ 7000000
v2 ^1.38 = .0262 <<< what now?
 
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If ##x^n=y##, what is x equal to? We learned how to do this in Intermediate Algebra in high school.
 
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Chestermiller said:
If ##x^n=y##, what is x equal to? We learned how to do this in Intermediate Algebra in high school.

I guess you are trying to trigger log out of me. That means, other things are correct though... :)
 
es4 said:
I guess you are trying to trigger log out of me. That means, other things are correct though... :)
I wouldn't use log. $$x=y^{1/n}$$
 
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Chestermiller said:
I wouldn't use log. $$x=y^{1/n}$$

hmm, seems like I got to brush up some old memories. Appreciated the response though.
 
es4 said:
hmm, seems like I got to brush up some old memories. Appreciated the response though.
Yes, I agree. In doing thermodynamics, the math is supposed to be a "gimme"
 
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Isn't ##~x^n = y~\Rightarrow~x = y^{1/n}~##taking the nth root?
Why wouldn't a log do, or an ln?
 
kumusta said:
Isn't ##~x^n = y~\Rightarrow~x = y^{1/n}~##taking the nth root?
Why wouldn't a log do, or an ln?
It’s just easier to calculate the value of 1/n and then raise y to that power.
 
I think you're right there. Just two steps away from the answer.
Using ln, just doing two computations won't give you the answer yet. You'll need to do at least three; first the value of ln##~y~##, then the value of (ln##~y##)/##n~##, then that of ##exp[{\frac {\ln y}{n}}]~## before you finally get ##x~.##
 
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