# Isothermal compression of ideal gas

1. Oct 22, 2007

### VooDoo

1. The problem statement, all variables and given/known data

I got this formula out of our textbook
$$W_{b}$$=($$P_{1}$$$$V_{1}$$)$$\textbf{ln}$$$$V_{2}/V_{1}$$
Because Isothermal:

$$P_{1}$$$$V_{1}$$=$$P_{2}$$$$V_{2}$$

Therefore the first equation can be written as:
$$W_{b}$$=($$P_{1}$$$$V_{1}$$)$$\textbf{ln}$$$$P_{1}/P_{2}$$

From this equation I worked the boundary work to be: 1364.4kJ. Using the following values: m=3kg, R = 0.4882, P(1) = 1400kpa, T(1) = 353K, V(1) = 0.369288 (meters cubed).

Now I have boundary work to be 1364.4kJ and Q(in) = 495 (kJ/kg) x 3kg = 1485kJ

To find the change in enthalpy dont I just add these values? I am not sure what I have done wrong.