Calculation of pressure & volume both isothermally & adiabatically

  • Thread starter es4
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  • #1
es4
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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 = 732


1.
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?
 

Answers and Replies

  • #2
22,322
5,201
If ##x^n=y##, what is x equal to? We learned how to do this in Intermediate Algebra in high school.
 
  • #3
es4
6
3
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... :)
 
  • #4
22,322
5,201
I guess you are trying to trigger log out of me. That means, other things are correct though... :)
I wouldn't use log. $$x=y^{1/n}$$
 
  • #5
es4
6
3
I wouldn't use log. $$x=y^{1/n}$$
hmm, seems like I got to brush up some old memories. Appreciated the response though.
 
  • #6
22,322
5,201
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"
 
  • #7
kumusta
76
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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?
 
  • #8
22,322
5,201
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.
 
  • #9
kumusta
76
26
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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