Calculating Work for Compressing Helium Gas at Constant Pressure and Temperature

[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, let's 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

 

[Spectre5]

assuming ideal gas of course :)

 

[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.