# Re-arranging an equation.

• Dec 20th 2010, 08:59 AM
DanBrown100
Re-arranging an equation.
Need help with a question guys, any help will be most appreciated.

The pressure, p, and volume, V, of a gas undergoing a polytropic process are related by the equation:

p1V1n = p2V2n

where n is the polytropic index.

If p1V1/ T1 = p2V2/ T2

determine and expression for p1 in terms of p2 , T1, T2 and n

It has got me a bit stumped guys so any help would be most appreciated.

Thanks
• Dec 20th 2010, 11:06 AM
emakarov
Quote:

The pressure, p, and volume, V, of a gas undergoing a polytropic process are related by the equation:

p1V1n = p2V2n

where n is the polytropic index.

Ifp1V1/ T1 = p2V2/ T2

determine and expression for p1 in terms of p2 , T1, T2 and n
From the last equation, $\displaystyle \displaystyle p_1=p_2\frac{T_1}{T_2}\frac{V_2}{V_1}$, so we have to find $\displaystyle \displaystyle\frac{V_2}{V_1}$. From the first equation, $\displaystyle \displaystyle\left(\frac{V_2}{V_1}\right)^n=\frac{ p_1}{p_2}$. By taking root of degree n of both sides, you can find $\displaystyle V_2/V_1$.

A remark concerning notation. As is standard, use * for multiplication (it can be omitted in certain cases like 3x) and ^ for exponentiation. Otherwise, it is not clear if your p1V1n means $\displaystyle p_1V_1n$ or $\displaystyle p_1V_1^n$.
• Dec 20th 2010, 01:28 PM
bjhopper
Dear Dan Brown,
I assume you have a chemistry problem and cannot get into the Chemistry Forum. If you describe your problem I will try to help

bjh
• Feb 11th 2011, 02:18 AM
dt71
so does V2/V1=(p1/p2)^1/n

and

p1 = p2T1/T2*(p1/p2)^1/n
• Feb 12th 2011, 10:59 AM
emakarov
Yes.
• Feb 12th 2011, 06:10 PM
dt71
Therefore
p1 = (p2T1p1^1/n)/T2p2^1/n

If I divide by p1^1/n

P1/P1^1/n = p2T1/T2p2^1/n

This is were i'm stuck
• Feb 13th 2011, 06:34 AM
emakarov
Quote:

This is were i'm stuck
What are you trying to do? The original problem was to "determine an expression for p1 in terms of p2 , T1, T2 and n". This has been done in post #4.
• Feb 13th 2011, 06:45 AM
dt71
but i still have p1 on both sides of the equation, i need to make p1 the subject
• Feb 13th 2011, 11:27 AM
emakarov
Sorry, you are right, of course.

So, we have $\displaystyle \displaystyle\frac{p_1}{p_1^{1/n}}=\frac{p_2}{p_2^{1/n}}\frac{T_1}{T_2}$, i.e., $\displaystyle \displaystyle p_1^{(n-1)/n}=p_2^{(n-1)/n)}\frac{T_1}{T_2}$. Raising both sides to n/(n-1), we get $\displaystyle \displaystyle p_1=p_2\left(\frac{T_1}{T_2}\right)^{n/(n-1)}$.
• Feb 13th 2011, 07:25 PM
bjhopper
rearanging an equation
PV^n = constant is used to calculate one unknown.polytropic process n=1.395
P1V1/T1 = P2V2/T2 is used to calculate a temperature
Example V1=12.48 cu ft V2=5.0 cu ft V1@ 70 F
P1 = 14.7 psia P2 = 52.6psia t=301F

bjh