# Thermodynamic Problem involving work

1. Feb 10, 2010

### Tomiscool

1. The problem statement, all variables and given/known data
A monatomic ideal gas is compressed adiabatically from a pressure of 1.19e5 Pa and volume of 250 L to a volume of 32.0 L.

(a) What is the new pressure of the gas in Pa?

(b) How much work is done on the gas?

2. Relevant equations
I know that W= Average P (Delta V)

3. The attempt at a solution
I used P (v^(5/3)) = P (V^(5/3)) to solve for P2, and i got 3.6e6 Pa which was correct
Now I just need help in solving for part b, the work done on the gas. I tried finding the average pressure between the two pressures and multiplying it by the change in volume, but Webassign said it was an incorrect answer. Any help please?

2. Feb 10, 2010

### cepheid

Staff Emeritus
Hi, welcome to PF!

What result did you get for the average pressure?

Also, your volumes are given in litres. Did you remember to convert everything to SI units? If you didn't, the answer you got for work would not be expressed in joules, which is why its numeric value would be incorrect.

3. Feb 10, 2010

### Andrew Mason

Since it is adiabatic, what can you say about the relationship between Work and change in internal energy? Can you find the change in internal energy of the gas? Use:

$$T_1V^{\gamma-1}_1 = T_2V^{\gamma-1}_2$$ and

$$\Delta U = nC_v\Delta T[/itex] AM 4. Feb 10, 2010 ### Tomiscool For the average, i got 1.889e6 Pascals. Ok, so converting the liters to m^3 i got .25 m^3and .032 m^3. RE Andrew Mason: thank you for the formulas, could you just explain the first one to me, and how I would go about using these formulas to solve for the Work. 5. Feb 10, 2010 ### cepheid Staff Emeritus Hey, just curious, (since you seem to know what you're talking about). Is the "average pressure" method a dead end for adiabatic compression (at least for the OP) because a linear change in volume won't result in a linear change in pressure, meaning that one won't be able to find the average P without integrating? 6. Feb 10, 2010 ### Tomiscool I know that there is no heat flow in, so the work must remain the same. Therefore the internal energy must remain the same correct? 7. Feb 10, 2010 ### Tomiscool Any help? 8. Feb 10, 2010 ### Andrew Mason NO! Apply the first law: [tex]\Delta Q = \Delta U + W$$ where W is the work done by the gas.

If $\Delta Q = 0$ what can you say about the relationship between W and $\Delta U$?

The adiabatic condition that you have used for the first part:

$$PV^{\gamma} = K$$ can be rewritten as:

$$(nRT/V)V^{\gamma} = nRTV^{\gamma-1} = K$$

You can use this to determine the change in temperature in an adiabatic process. You can use that change in temperature to determine the change in internal energy and, hence, the work done in the adiabatic process.

AM