Thermodynamics - Carnot Engine

[Wellesley]

Homework Statement

A Carnot engine has 40 g of helium gas as its working substance. It operates between two reservoirs of temperatures T1 = 400°C and T2 = 60°C. The initial volume of the gas is V1 = 2.0 X 10-2 m3, the volume after isothermal expansion is V2 = 4.0 X 10-2 m3, and the volume after the subsequent adiabatic expansion is V3 = 5.0 X 10-2 m3.

(a) Calculate the work generated by this engine during one cycle.

(b) Calculate the heat absorbed and the waste heat ejected during one cycle.

Homework Equations

The Attempt at a Solution

I'm not sure how tor eally start this problem...

I calculated the effiency of the carnot engine via this:

1-333K/673K-->0.505201

At this point, I wasn't sure how to treat an isothermal expansion. I know that delta U=0, and delta Q= delta W.

I'm not looking for someone to do this for me. I would really appreciate it though, if someone could give me a clue where to start. Thanks.

 

[rock.freak667]

The Attempt at a Solution

I'm not sure how tor eally start this problem...

I calculated the effiency of the carnot engine via this:

1-333K/673K-->0.505201

At this point, I wasn't sure how to treat an isothermal expansion. I know that delta U=0, and delta Q= delta W.

You should start by finding the P,V at all the four states. You can easily find P₁ using the gas equation.

They tell you that it expands isothermally, so that means

PV=constants or P₁V₁=P₂V₂

You can now find P₂.

For adiabatic expansion PV^γ=constant. You will need to get γ for helium.

Then it just becomes a case of using Q-W=ΔU

 

[Andrew Mason]

You have worked out the efficiency. All you need to know is Qh. - the heatflow INTO the gas. This occurs only during isothermal expansion. Work out the Qh and you will solve this problem. (Hint: apply the first law to determine Qh).

AM

 

[Wellesley]

Thanks for the replies...

Andrew Mason, your response really helped. I tried rock.freak667's solution first, but I my pressure values were off.

Then I tried an equation that was buried in the chapter : Q_h=nRT₁ * ln (V₂/V₁), which was exactly what I needed. The rest was a piece of cake.

Thanks again for both of your replies.

 

[Andrew Mason]

Thanks for the replies...

Andrew Mason, your response really helped. I tried rock.freak667's solution first, but I my pressure values were off.

Then I tried an equation that was buried in the chapter : Q_h=nRT₁ * ln (V₂/V₁), which was exactly what I needed. The rest was a piece of cake.

Thanks again for both of your replies.

For an isothermal process, there is no change in internal energy. So $\Delta Q = W$

So to find the heat flow into the gas, determine the work done at constant temperature (substituting P = nRT/V):

$$W = \int_{V_i}^{V_f} Pdv = nRT\int_{V_i}^{V_f} dV/V = nRT\ln\left(\frac{V_f}{V_i}\right)$$

That is where your equation comes from.

AM