## Another Thermo Question

1. The problem statement, all variables and given/known data

See Attachment

3. The attempt at a solution

Well to start off I know the gas is monatomic

I can find the work done by finding the area under each curve/line:

$$W_{AB} = p_0(2v_0-v_0) = p_0v_0$$
$$W_{CD} = (p_0/32)(8v_0-16v_0) = \frac{-p_0v_0}{4}$$

I cant think of how to find the area under the other two curves, im guessing integration but I dont know how to set it up.

After I find these other to, what do I do?

Thanks
Attached Thumbnails

 PhysOrg.com science news on PhysOrg.com >> Heat-related deaths in Manhattan projected to rise>> Dire outlook despite global warming 'pause': study>> Sea level influenced tropical climate during the last ice age
 Recognitions: Science Advisor I can't see the attachment yet, but finding the area between two curves is easy. The easiest way to do it is to subtract the area under the lower curve from the area under the upper curve. If you did it in one integral, you could set up the limits of integration from one curve to the other.

## Another Thermo Question

for isentropic expansion and compression in ICEs we use the same principle u used to find the first 2 works.
we use : W= (PoVo-PoVo/4)/(1-k) k=Cp/Cv. for isentropic adiabatic compression or expansion.

How did u know the gas was monoatomic?

Recognitions:
Homework Help
 Quote by eaboujaoudeh for isentropic expansion and compression in ICEs we use the same principle u used to find the first 2 works. we use : W= (PoVo-PoVo/4)/(1-k) k=Cp/Cv. for isentropic adiabatic compression or expansion.
Where do you get this? It works for BC but not DA

(1) $$W = K\frac{V_f^{1-\gamma} - V_i^{1-\gamma}}{1-\gamma}$$

where $K = PV^\gamma$

This is just the integral $\int dW$ where $dW = dU = PdV = KV^{-\gamma}dV$ (dQ=0)

Since for DA $P_f = 32P_i$ and $V_f = V_i/8$ the numerator in (1) is simply:

$$P_fV_f - P_iV_i = 32P_iV_i/8 - P_iV_i = 3P_iV_i$$

for BC, $P_f = P_i/32$ and $V_f = 8V_i$ the numerator in (1) is simply:

$$P_fV_f - P_iV_i = 8P_iV_i/32 - P_iV_i = -3P_iV_i/4$$

 How did u know the gas was monoatomic?
Apply the adiabatic condition $PV^\gamma = constant$ to one of the adiabatic paths and solve for $\gamma$.

AM

 Thanks Andrew