Calculating Heat Added/Removed to Ideal Gas System

[mikefitz]

Homework Statement

A monatomic, ideal gas is in a sealed container (the number of gas molecules is always constant: n = 2 moles); the initial pressure is Pi = 1.01 x 10^5 Pa and the initial volume is Vi = 0.0224 m^3.

* First, the volume of the gas is decreased at a constant pressure (at Pi = 1.01 x 10^5 Pa) to a final volume of Vf = 0.0155 m^3.
* Second, the pressure of the gas is increased at a constant volume (at Vf = 0.0155 m3) to a final pressure of Pf = 1.35 x 10^5 Pa.

How much heat was added to (give as a positive number) or removed from (give as a negative number) the system? (The gas constant R = 8.31 J/mole-K.)

Homework Equations

PV = nRT

The Attempt at a Solution

I guess I am confused as to how I am supposed to solve this problem without knowing the heat capacity or the specific heat of the substance.

I have calculated Ti=94.14 and Tf=125.83 - deltaT=31.692 C - why is this incorrect?

 

[courtrigrad]

The heat capacity at constant volume of an ideal gas is: c_{v}NR.

c_{v} = \frac{3}{2} for a monatomic gas, and \frac{5}{2} for a diatomic gas.

The heat capacity at constant pressure of an ideal gas is:

(c_{v}+1})NR

 

[mikefitz]

Problem Solved.

Please look at case one of the pressure vs volume graph:

http://img99.imageshack.us/img99/2964/followuphz6.gif

I thought to get the total work done by the system I would take the work done by decreasing volume and add the work done by increasing pressure.

To get the total work done all I had to do was use the work done by the system decreasing volume. (-696.9J)

Why didn't I have to add on the work done by the increase in pressure? I calculated that to be 527J, where does this energy disappear to?

thanks

 

[courtrigrad]

The work done is the area under the graph of pressure versus volume. When you increase the volume, the gas does positive work.

W = P\Delta V

For non-constant pressure the work is:

W = \int_{V_{1}}^{V_{2}} P\; dV