# Homework Help: Check my work on work please

1. Nov 14, 2004

### Spectre5

I just want to make sure I am doing this correctly...

a 2000 cm^3 container holds 0.10 mol of heliuum gas at 300 C. How much work must be done to compress the gas to 1000 cm^3 at
a) constant pressure
b) constant temperature

So...
2000 cm^3 = .002 m^3
1000 cm^3 = .001 m^3
300 Celsius = 573 K

$$W=-\int_{V_1}^{v_2}{pdv}$$

where W is work, v is volume and p is pressure....this is the work that the environment does on the system (that is why the negative sign is in front...I know that most books present the work the gas does on the environment, but this book is a little weird I guess)

Furthermore, lets use the ideal gas law to calculate the initial pressure..

$$PV=nRT$$
$$P(.002)=(.1)(8.31)(573)$$
$$P=238.082 KPa$$

So...
a) Constant pressure

$$W=-\int_{V_1}^{v_2}{pdv}$$

$$W=-\int_{.002}^{.001}{238.082\times 10^3 dv}$$

$$W=238 J$$

b) Constant temperature

$$W=-\int_{V_1}^{v_2}{pdv}$$

$$W=-\int_{V_1}^{v_2}{(\frac{nRT}{V})dv}$$

$$W=-\int_{.002}^{.001}{(\frac{(.1)(8.31)(573)}{V})dv}$$

$$W=330 J$$

Do these seem correct?

Thanks for any input..it is much appreciated

2. Nov 14, 2004

### Spectre5

assuming ideal gas of course :)

3. Nov 14, 2004

### CartoonKid

Your work are correct. The book is not weird but rather tricky. In fact, in the exam, we do have to be careful of how the question is being set.