How Is Energy Managed in a Cyclic Process of an Ideal Diatomic Gas?

[pentazoid]

Homework Statement

An ideal diatomic gas, in a cylinder with a movable piston, undergoes the rectangular cyclic process. Assume that the temperature is always such that rotational degrees of freedom are active, but vibrational modes are "frozen out". Also Assume that the only type the only type of work done on the gas is quasistatic compression-expansion work.

For each of the four steeps A through D , compute the work on the work done on the gas, the heat added to the gas, and the change in the energy content of the gas. Express all answers in the terms P_1, P_2, V_1, and V_2

Steps A, B, C and D are drawn on a PV diagram

State A: Pressure changes, Volume is constant(V_1 is constant)

State B: Volume changes, Pressure is constat(P_2 is constant)

State C: Pressure changes , Volume is constant)(V_2 is constant)

State D: Volume changes, pressure is constant(P_1 is constant)

Homework Equations

delta(U)=W+Q

delta(U)=f/2*NkT (equipartition thm. for energy)

NkT=PV (ideal gas law)

W=-integral(from V_initial to V_final) P(V) dV (quasistatic)

W= -P*delta(V) (quasistatic)

The Attempt at a Solution

W_A=0
W_B=-P_2(V_2-V_1)=
W_C=0
W_A=-P_1(V_1-V_2)=P_1(V_2-V_1)

U=f/2*N*k*delta(T)=f/2*P*delta(V)
since NkT=PV
f=5 since air is a polyatomic molecule

U_A=5/2*(P_2*V_1-P_1*V_1)=5/2*(V_1)*(P_2-P_1)
U_B=5/2*(P_2*V_2-P_2*V_1)=5/2*(P_2)*(V_2-V_1)
U_C=5/2*(P_1*V_2-P_2*V_2)=5/2*(V_2)*(P_1-P_2)
U_D=5/2*(P_1*V_1-P_1*V_2)=5/2*(P_1)*(V_1-V_2)

to calculate the heat, you would use the equation delta(U)=Q+W,Q=delta(U)-W for each of the four states.

 

[pentazoid]

Did anyone not understand my question?