Closed cycle of an ideal gas

[imdesperate]

Homework Statement

An ideal gas (degree of freedom = 5) is in a closed cycle like the given figure. Find the efficiency of the cycle.

Relevant Equations

$\Delta U = A + Q$
$\eta = \frac{-A}{Q_+}$

Hello PF, this is my first time posting here. I will try my best to make my formulas readable.

So I know what needed to be done:
The efficiency is calculated by the formula: $\eta = \frac{-A}{Q_+}$
With $A$ being the total work done in the cycle, $Q_+$ being the heat absorbed in the cycle. The opposite of $Q_+$ is $Q_-$ being the heat released.

Heat ratio $\gamma = 1 + \frac2 i = 1 + \frac2 5 = \frac7 5$ (notated degree of freedom as i).

First, I calculated the pressure and volume of each state using the relation of each process.
For process (1-2): isothermal, I calculated $p_2$ using
$p_1 * V_1 = p_2 * V_2$
$p_2 = \frac5 3 * 10^5 Pa$
For process (2-3): adiabatic, I calculated $V_3$ using
$p_2 * (V_2)^\gamma = p_3 * (V_3)^\gamma$
$V_3 = 12.96 * 10^{-3} m^3$
For process (3-4): isobaric, so $p_3 = p_4 = 10^5 Pa$.
For process (4-1): isochoric, so $p_4 = p_1= 3* 10^{-3} m^3$.

Then, I can calculate the work done for each process.

(1-2): isothermal, $A_{12} = nRT\ln(\frac{V_1}{V_2}) = pV\ln(\frac{V_1}{V_2}) = -1647.92 J$
(2-3): adiabatic, $A_{23} = \frac{p_2 * V_2 - p_3 * V_3} {\gamma - 1} = 509.25 J$
(3-4): isobaric, $A_{34} = p * (V_3 - V_4) = 996.3 J$
(4-1): isochoric, $A_{41} = 0 J$
So $A = A_{12} + A_{23} + A_{34} + A_{41} = -142.37 J$

I proceeded to calculate the heat absorbed or released for each process.

(1-2): isothermal, $Q_{12} = -A_{12} = 1647.92 J > 0$ so $Q_{12}$ is $Q_{+12}$
(2-3): adiabatic, $Q_{23} = 0 J$
(3-4): isobaric, $Q_{34} = n * C_p * (T_4 - T_3)$. with $C_p = (\frac i 2 + 1)R$ ($R$ is the gas constant)
(4-1): isochoric, $Q_{41} = n* C_v *(T_1 - T_4)$, with $C_v = \frac{iR}{2}$

The problem starts here, I was not given the moles for the gas (I have confirmed this with my teacher).

I tried to calculate the temperature T for each state. By using the relation of each process, I managed to figure out the relation between the temperature: $$T_1 = T_2 = \frac{5T_3}{\frac{V3}{V4}} = 5T_4$$.

I've gotten these through

(1-2): isothermal, $T_1 = T_2$;
(3-4): isobaric, $T_3 = (\frac{V_3}{V_4})T_4$;
(4-1): isochoric, $T_1 = 5T_4$.

I've used using the formula $\Delta U = A + Q$ to try and find any kind of equation to calculate the temperature, but I've gotten nowhere, all the results I got are the same relation I'd gotten from earlier $(T_1 = T_2 = \frac{5T_3}{\frac{V3}{V4}} = 5T_4)$

I would like a hint on where to start!

 

[kuruman]

You don't need the number of moles to calculate the efficiency. The efficiency is a ratio with $n$ in the numerator and the denominator which will eventually cancel. First express all pressures and volumes in terms of $p_1$ and $V_1$.

A convenient way to summarize your results would be to make a Table as show below and fill in the entries. Use symbols instead of numbers. You can always substitute numbers at the very end. I already put in the zero that you found, there are more. Be sure that the totals are what they should be over a complete cycle. The total in column $W$ divided by the sum of the positive entries in column $Q$ is the efficiency.

	$$\Delta U$$	$$Q$$	$$W$$
$$1\rightarrow 2$$
$$2\rightarrow 3$$
$$3\rightarrow 4$$
$$4\rightarrow 1$$			$$0$$
$$\rm{TOTAL}$$

 

[imdesperate]

Well I did it.
I used $T_1 = 5T_4$ to figure out $Q_{41}$ is positive. Then, I put them all in the formula and converted all pressures and volumes into $p_1$ and $V_1$. I also converted all the temperatures into the form $\frac{pV}{nR}$ which made $nR$ cancel out. Finally I got $0.0306$ as the answer.

Thank you so much!

 

[kuruman]

Well I did it.
I used $T_1 = 5T_4$ to figure out $Q_{41}$ is positive. Then, I put them all in the formula and converted all pressures and volumes into $p_1$ and $V_1$. I also converted all the temperatures into the form $\frac{pV}{nR}$ which made $nR$ cancel out. Finally I got $0.0306$ as the answer.

Thank you so much!

I got a higher value or the efficiency by an order of magnitude. You don't show the numbers that went in your calculation. If you wish to pursue this further, please show your efficiency calculation so that we can pinpoint the problem with it. Remember, the efficiency is the total work done by the gas divided by the total amount of heat that enters the gas.