The work done by a perfect gas

[Emspak]

Homework Statement

An ideal gas at initial temperature T₁ and pressure P₁ is compressed by a piston to half its original volume. The temperature is varied so that the relation P=AV always holds and A is a constant. What is the work done n the gas in terms of n (moles of gas) R (gas constant) and T₁?

Homework Equations

Using ideal gas law, PV= nRT

and assuming the work done is [itex]W = \int^{V_b}_{V_a} P dV[/itex]

The Attempt at a Solution

OK, simple enough, right?

V_b = (1/2) V_a

So V = nRT/P and [itex]V_a = \frac{nRT}{P_1}[/itex] and [itex]V_b = \frac{nRT}{2P_1}[/itex]

Plug this into the integral above. Since P=AV (in the givens) it should be

[itex]W = \int^{V_b}_{V_a} (AV) dV = A \int^{V_b}_{V_a} V dV = A\frac{V^2}{2}|^{V_a}_{V_b}[/itex]

Which leaves me with

[itex]=\frac{A}{2} \left[\left(\frac{nRT}{2P_1}\right)^2 - \left(\frac{nRT}{P_1}\right)^2\right] = \frac{A}{2}\left(\frac{nRT}{P_1}\right)^2(-3/4) = \frac{-3A}{8}\left(\frac{nRT}{P_1}\right)^2[/itex]

Yet the answer is listed as [itex]\frac{-3A}{8}\left(\frac{nRT}{P_1}\right)[/itex]

So I am trying to figure out how I got the extra factor in there. I think I did everything right, but is there some stupid mathematical error I made?

Anyhow, it's possible there's a typo in the book's answer too. But...

any help is appreciated, even though I anticipate it will be trivial...

 

[Andrew Mason]

Homework Statement

An ideal gas at initial temperature T₁ and pressure P₁ is compressed by a piston to half its original volume. The temperature is varied so that the relation P=AV always holds and A is a constant. What is the work done n the gas in terms of n (moles of gas) R (gas constant) and T₁?

Homework Equations

Using ideal gas law, PV= nRT

and assuming the work done is [itex]W = \int^{V_b}_{V_a} P dV[/itex]

The Attempt at a Solution

OK, simple enough, right?

V_b = (1/2) V_a

So V = nRT/P and [itex]V_a = \frac{nRT}{P_1}[/itex] and [itex]V_b = \frac{nRT}{2P_1}[/itex]

Plug this into the integral above. Since P=AV (in the givens) it should be

[itex]W = \int^{V_b}_{V_a} (AV) dV = A \int^{V_b}_{V_a} V dV = A\frac{V^2}{2}|^{V_a}_{V_b}[/itex]

Which leaves me with

[itex]=\frac{A}{2} \left[\left(\frac{nRT}{2P_1}\right)^2 - \left(\frac{nRT}{P_1}\right)^2\right] = \frac{A}{2}\left(\frac{nRT}{P_1}\right)^2(-3/4) = \frac{-3A}{8}\left(\frac{nRT}{P_1}\right)^2[/itex]

Yet the answer is listed as [itex]\frac{-3A}{8}\left(\frac{nRT}{P_1}\right)[/itex]

So I am trying to figure out how I got the extra factor in there. I think I did everything right, but is there some stupid mathematical error I made?

Anyhow, it's possible there's a typo in the book's answer too. But...

any help is appreciated, even though I anticipate it will be trivial...

The correct answer is 3nRT₁/8, which can be derived from your answer.

The given answer is not dimensionally correct. The answer has to have dimensions of energy (work) ie. PV or nRT. Since A has dimensions of P/V, the correct answer has to have dimensions of AV². The given answer has dimensions of AV = P.AM

 

[Emspak]

Thanks lots, I knew there was something stupid I was missing. I looked at it again and the given answer was yours, but I forgot about just plugging P/V in for A. D'OH!