Thermodynamic cyclic process - 2 isobaric and 2 adiabatic

[Thermofox]

Homework Statement

A heat engine that uses an ideal gas carries out a cycle composed by:
$A\rightarrow B$ isobaric expansion
$B\rightarrow C$ adiabatic expansion
$C\rightarrow D$ isobaric compression
$D\rightarrow A$ adiabatic compression
All processes are reversible. I know that $P_{AB}= 20 atm$ ; $P_{CD}= 10 atm$ and that $V_A= 5 L$ ; $V_B= 10 L$.

Assuming that the heat engine is able to work with a mole of a monoatomic and a mole of a diatomic gas determine:
1) The total heat absorbed by the gasses in a cycle
2) The minimum temperature reached by the gasses in a cycle
3) $W_{net}^{mono}$ and $W_{net}^{bi}$
4) The efficiency of the heat engine for both gasses.

Relevant Equations

$\Delta U = W + Q$

I think I know how to solve the problem, but I've incurred into some problems with my computations.
Let's take for example the first question, in the monoatomic case.

The total heat absorbed by the gas coincides with the heat absorbed during $A\rightarrow B$.
Since the pressure is constant $Q_{AB}= nc_p \Delta T$. To find the heat I need to first find $T_A$ and $T_B$. That's not a problem since The gas is ideal hence: $$T_A= \frac {P_A V_A} {nR}= \frac {(20) (5)} {(1)0.0821}= 1218K ; T_B= \frac {(20) (10)} {(1)0.0821}= 2436K $$
Therefore $Q_{AB}= nc_p \Delta T= \frac 5 2 8.31 (2436 - 1218)= 25304J$.

The problem is that if I find $\Delta U_{AB} = nc_v\Delta T= \frac 3 2 8.31 (2436-1218)\approx 15182J$, now $\Delta U = Q + W$ but that isn't the case since $W_{AB}= -P_{AB} \Delta V= -20(10-5)= -100J$
$\Rightarrow 25304J -100J \neq 15182J$. I don't know where I'm making a mistake, but there must be one.

 

[robphy]

It might help to include units in your calculation.
$(20{\rm\ atm})(10 {\rm\ L} - 5 {\rm\ L})=100{\rm\ atm\cdot L}$

 

[kuruman]

What are your units of pressure and volume in your work calculation?

The numbers for the temperatures come out correctly when pressures in atm are converted to Pa, volumes in L converted to m^-3 and 0.0821 (R) in the denominator is converted to 8.31 J/(mol K).

What I don't understand is if we have a heat engine with 1 mole of monatomic and 1 mole of diatomic gas mixed together or two separate heat engines with one mole of gas each.

It might help to include units in your calculation.

I think OP should do the calculations symbolically and then substitute. It will ease considerably the task of tracing what is going on and offer help.

 

[Thermofox]

two separate heat engines with one mole of gas each.

This.

It might help to include units in your calculation.
$(20{\rm\ atm})(10 {\rm\ L} - 5 {\rm\ L})=100{\rm\ atm\cdot L}$

The numbers for the temperatures come out correctly when pressures in atm are converted to Pa, volumes in L converted to m-3 and 0.0821 (R) in the denominator is converted to 8.31 J/(mol K).

That's the mistake, I treated $L\space atm$ as $J$

When I did $W_{AB}= -P_{AB} (V_B - V_A)= -20atm(10-5)L\neq -100J$, but $-100 atm\space L=101,325J(-100)\approx -10,132J$

Now $\Delta U_{AB} = Q + W= 25,304J -10,132J = 15,172J$.

If I compare it with $\Delta U= nc_v\Delta T_{AB}= 15,182J$, there is a difference of $10J$ but it is normal considering the approximations I did.

 

[kuruman]

The problem is that if I find $\Delta U_{AB} = nc_v\Delta T= \frac 3 2 8.31 (2436-1218)\approx 15182J$, now $\Delta U = Q + W$ but that isn't the case since $W_{AB}= -P_{AB} \Delta V= -20(10-5)= -100J$
$\Rightarrow 25304J -100J \neq 15182J$. I don't know where I'm making a mistake, but there must be one.

The problem is that you are substituting numbers when you should be using symbols.
For A→B
Note that since the volume doubles at constant pressure, $T_B=2T_A$
Then
$\Delta U =\frac{3}{2}nR(2T_A-T_A)=\frac{3}{2}nRT_A=\frac{3}{2}p_AV_A.$
$Q=\frac{5}{2}nR(2T_A-T_A)=\frac{5}{2}nRT_A=\frac{5}{2}p_AV_A.$
$W=-p_AV_A.$
Clearly the first law is satisfied,$$\Delta U=Q+W\rightarrow\frac{3}{2}p_AV_A=\frac{5}{2}p_AV_A-p_AV_A.$$Now you substitute the given values for $V_A$ and $p_A.$ Be sure to convert to Pascals for pressure and cubic meters for volume.

 

[Thermofox]

The problem is that you are substituting numbers when you should be using symbols.
For A→B
Note that since the volume doubles at constant pressure, $T_B=2T_A$
Then
$\Delta U =\frac{3}{2}nR(2T_A-T_A)=\frac{3}{2}nRT_A=\frac{3}{2}p_AV_A.$
$Q=\frac{5}{2}nR(2T_A-T_A)=\frac{5}{2}nRT_A=\frac{5}{2}p_AV_A.$
$W=-p_AV_A.$
Clearly the first law is satisfied,$$\Delta U=Q+W\rightarrow\frac{3}{2}p_AV_A=\frac{5}{2}p_AV_A-p_AV_A.$$Now you substitute the given values for $V_A$ and $p_A.$

Okay, I was able to finish the problem. Thanks for the help!