# What Is the Work Done by a Perfect Gas During Compression?

• Emspak
In summary, the conversation discusses finding the work done by an ideal gas at initial temperature T1 and pressure P1 when compressed by a piston to half its original volume. Using the ideal gas law and assuming the work done is equal to the integral of pressure over volume, the work is calculated to be -3nRT1/8. However, the given answer is incorrect as it does not have the correct dimensions of energy. By plugging in P/V for A, the correct answer can be derived as 3nRT1/8.
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 $W = \int^{V_b}_{V_a} P dV$

## The Attempt at a Solution

OK, simple enough, right?

Vb = (1/2) Va

So V = nRT/P and $V_a = \frac{nRT}{P_1}$ and $V_b = \frac{nRT}{2P_1}$

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

$W = \int^{V_b}_{V_a} (AV) dV = A \int^{V_b}_{V_a} V dV = A\frac{V^2}{2}|^{V_a}_{V_b}$

Which leaves me with

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

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

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...

Emspak said:

## 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 $W = \int^{V_b}_{V_a} P dV$

## The Attempt at a Solution

OK, simple enough, right?

Vb = (1/2) Va

So V = nRT/P and $V_a = \frac{nRT}{P_1}$ and $V_b = \frac{nRT}{2P_1}$

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

$W = \int^{V_b}_{V_a} (AV) dV = A \int^{V_b}_{V_a} V dV = A\frac{V^2}{2}|^{V_a}_{V_b}$

Which leaves me with

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

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

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 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

1 person
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!

## 1. What is the definition of work done by a perfect gas?

The work done by a perfect gas is the product of the pressure and change in volume of the gas. It is the amount of energy transferred to or from the gas due to a change in its volume.

## 2. How is the work done by a perfect gas calculated?

The work done by a perfect gas is calculated using the formula W = PΔV, where W is work, P is pressure, and ΔV is the change in volume of the gas.

## 3. Can the work done by a perfect gas be negative?

Yes, the work done by a perfect gas can be negative. This occurs when the gas expands and does work on its surroundings, resulting in a decrease in its internal energy.

## 4. What is the SI unit for work done by a perfect gas?

The SI unit for work done by a perfect gas is joule (J). It can also be expressed in other units such as newton-meter (N·m) or pascal-cubic meter (Pa·m3).

## 5. How is the work done by a perfect gas related to the first law of thermodynamics?

The first law of thermodynamics states that the change in internal energy of a system is equal to the heat transferred to or from the system, minus the work done by the system. Therefore, the work done by a perfect gas is directly related to the change in its internal energy and can be used to determine the heat transferred to or from the gas.

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