Calculating temperatures of reservoirs in Carnot engine

[arenaninja]

Homework Statement

An ideal monatomic gas is the working substance of a Carnot engine. During the isothermal expansion, the volume doubles. The ratio of the final volume to the initial volume in the adiabatic expansion is 5.7. The work output of the engine is $$9x10^{5}J$$ in each cycle. Compute the temperature of the reservoirs between which the engine operates.

Homework Equations

$$W_{eng}=\left|Q_{h}\right| - \left|Q_{c}\right|$$
Since the engine is a Carnot engine, we know that Q=0 for the adiabatic compression and adiabatic expansion. We also know that for legs 1 and 3 (isotherms):
$$\left|W_{n}\right|=nRT ln\left(\frac{V_{f}}{V_{i}}\right)$$
and that Q=W for these isotherms.

From the equation for the work of the engine, we can derive a relation between $$T_{c}$$ and $$T_{h}$$. From there on, we can use
$$TV^{\gamma-1}=constant$$

The Attempt at a Solution

Ok. The initial volume of the system is V, and the problem states that at the end this will be equal to 2V. So the ratio of (Vf/Vi) = 2. Thus,
$$\left|Q_{1}\right|=nR ln \left(2\right)$$
And here's where my confusion begins. The problem states that the ratio of the final volume to the initial volume in the adiabatic expansion is 5.7. To me, this means that (Vf/Vi)=(1/5.7) (because after isothermal compression, the system should go back to the initial volume V, correct?). So,
$$\left|Q_{2}\right|=nR ln \left(1/5.7\right)$$

But I attempted this, and it simply didn't work. I used $$\gamma=\frac{5}{3}$$, which is correct for monatomic ideal gas.

Can anyone tell me what I am interpreting wrong? I know the lower temperature is in the range of ~70K, but I've tried a few things and I'm unable to find it.

Many thanks.

 

[quenderin]

There are two isothermal and two adiabatic processes in the Carnot cycle. After the isothermal compression, we have an adiabatic compression to return to the original state.