Thermodynamic Cycle: Calculating Heat Transfer and Work

[kdinser]

Sorry about the double post, I had technical difficulties that I was working on and the title of my orginal post got screwed up.

I'm having problems getting started on this one.

A gas undergoes a thermodynamic cycle consisting of 3 processes

process 1-2 compression with pressure(p)*volume(V) = constant, from
p_{1} = 1 bar
V_{1} = 1.6m^3
to
p_{2} = ?
V_{1} = .2m^3

U_{2}-U_{1}=0

process 2-3
Constant pressure to V_{3}=V_{1}

process 3-1
Constant Volume, U_{1}-U_{3} = -3549kJ

There are no significant changes in kinetic or potential energy.
Determine the heat transfer and work for process 2-3 in kJ.

I don't have any problems findingp_{2} or the work needed to compress the gas, but I'm not really sure where to go from there.

p_2=\frac{p_1V_1}{V_2}

W=\int p dV

When I work these out, I end up with 333kJ for W and 8 bar for p2.

If someone could give me a quick push in the right direction, that would be great.

 

[Andrew Mason]

A gas undergoes a thermodynamic cycle consisting of 3 processes

process 1-2 compression with pressure(p)*volume(V) = constant, from
p_{1} = 1 bar
V_{1} = 1.6m^3
to
p_{2} = ?
V_{1} = .2m^3

U_{2}-U_{1}=0

process 2-3
Constant pressure to V_{3}=V_{1}

process 3-1
Constant Volume, U_{1}-U_{3} = -3549kJ

There are no significant changes in kinetic or potential energy.
Determine the heat transfer and work for process 2-3 in kJ.
.

The work done between 2-3 is just P_2\Delta V = P_2(V_3-V_2) = P_2(7*V_2)

The change in internal energy is U_3-U_2 = U_3-U_1, since U_2=U_1

Use \Delta Q = \Delta U + W to find the energy (heat) flow into the system.

AM