Urgent! Max Heat Withdrawn from Low-Temp Reservoir

[Telemachus]

Urgent! Thermodynamic refrigerator

Homework Statement

One mole of a monatomic ideal gas is allowed to expand isothermally from an initial volume of 10 liters to a final volume of 15 liters, the temperature being maintained at 400K. The work delivered is used to drive a thermodynamic refrigerator operating between reservoirs of temperatures 200 and 300K. What is the maximum of heat withdrawn from the low-temperature reservoir?

This is what I did:
$$u=cRT \rightarrow T=\frac{cR}{u}$$

$$Pv=RT \rightarrow P=\frac{RT}{v}$$

$$u=cRT=constant$$

$$du=\frac{u}{cR}ds-\frac{RT}{v}dv=0 \rightarrow \Delta s=r \ln\frac{v_f}{v_0}$$

$$Q_{12}=T \Delta S=400R \ln 1.5=-W_{12}$$

The thing is I'm having a negative work. I assume I'm making something wrong, I just can't see the mistake.

I have my exam tomorrow, so any help will be appreciated.

 

[Andrew Mason]

Homework Statement

One mole of a monatomic ideal gas is allowed to expand isothermally from an initial volume of 10 liters to a final volume of 15 liters, the temperature being maintained at 400K. The work delivered is used to drive a thermodynamic refrigerator operating between reservoirs of temperatures 200 and 300K. What is the maximum of heat withdrawn from the low-temperature reservoir?

This is what I did:
$$u=cRT \rightarrow T=\frac{cR}{u}$$

$$Pv=RT \rightarrow P=\frac{RT}{v}$$

$$u=cRT=constant$$

$$du=\frac{u}{cR}ds-\frac{RT}{v}dv=0 \rightarrow \Delta s=r \ln\frac{v_f}{v_0}$$

$$Q_{12}=T \Delta S=400R \ln 1.5=-W_{12}$$

The thing is I'm having a negative work. I assume I'm making something wrong, I just can't see the mistake.

I have my exam tomorrow, so any help will be appreciated.

It is not that complicated. Calculating the work done by the gas is fairly straight forward and does not require calculation of the heat flow:

$$W = \int_{V_0}^{V_f}PdV = \int_{V_0}^{V_f}RTdV/V = RT\ln(V_f/V_0) = RT\ln(1.5)$$

If you set $\Delta Q = -W$ you are stating that W is the work done ON the gas. So in your result $W_{12}$ is the work done ON the gas.

AM

 

[Telemachus]

Thanks!

I have another problem now.
I've proceeded this way with the exercise. From then, I've made a scheme, similar to the exercise you helped me before (I think its pretty much like the same).

$$\ln \frac{T_{AF}}{T_{A0}}=\ln \frac{T_{BF}}{T_{0F}} \rightarrow T_{AF}=\frac{2T_{BF}}{3}$$

$$W=Q_h-Q_c \rightarrow 400\ln(1.5)=\frac{3}{2}(T_{BF}-300)-\frac{3}{2}(T_{AF}-200)\rightarrow T_{BF}\approx 624K$$

Now, I try to get Qc, but I think the result I get is wrong, I don't want you to make the numbers, just want you to tell me if what I did is wrong, or if its okey and the mistake is given by a simple error of calculus.

$$Q_c=W-Q_h \rightarrow Qc=-400R\ln(1.5)+R\frac{3}{2}(624-300)$$
It gives $$Q_c \approx R 324$$
I think it doesn't make much sense this result. But I'm not sure. The impression it gave me is that its bigger than the internal energy of the system A for it's initial state, which will be a big contradiction. U0=cRT=3/2 R 200K

Thank you so very very much for your predisposition for helping and your fast answer!