Thermodynamics imaginary heat machine?

[shahar weiss]

Homework Statement

given the following diagram : http://www.freeimagehosting.net/uploads/…

a)mention for every step if it is isothermal / adiabatic / else. does
the system receive heat or emit heat.
b)given this gas is ideal gas, sketch a diagram with respect to P and V
c)calculate work done by 1 cycle, also calculate heat received by the
machine and its efficiency, compare results to carnot engine between
500k and 300k temp diff.

Homework Equations

S/R = ds/R = dQ/RT
ideal gas eq. pv=Rt (n = cons. here)

The Attempt at a Solution

i did part a and found that the process is isothermal->adiabatic->isothermal->adiab…

i cannot decide though where is the expansion and where is compression so I am having hard time decide about and p and v relations.
i would be very happy for assistance

 

[rock.freak667]

The Attempt at a Solution

i did part a and found that the process is isothermal->adiabatic->isothermal->adiab…

i cannot decide though where is the expansion and where is compression so I am having hard time decide about and p and v relations.
i would be very happy for assistance

While I cannot find your diagram in that link, I have pretty good idea of how your diagram looks.

When you compress something, the pressure increases, so on your diagram when the pressure goes from low to high during an isothermal process, compression occurs.

During an isothermal process when the pressure goes from high to low, expansion occurs (since the volume of the gas will increase).

 

[shahar weiss]

the image is at
http://img547.imageshack.us/i/newbitmapimagez.jpg/

problem is i cannot decide when the system is comprassing and when its expanding

 

[rock.freak667]

the image is at
http://img547.imageshack.us/i/newbitmapimagez.jpg/

problem is i cannot decide when the system is compressing and when its expanding

You need to follow the arrows. From point 'f' to 'a', the pressure is increasing so that the final temperature is higher than the initial temperature, so compression is occurring there.

 

[shahar weiss]

i just thought of somehting else, i can know what's the entropy change in the diagram from point to point while knowing the temp diff in those 2 points. maybe this gives me the info of
whether its expansion or comprassion

 

[shahar weiss]

You need to follow the arrows. From point 'f' to 'a', the pressure is increasing so that the final temperature is higher than the initial temperature, so compression is occurring there.

how can you know pressure is increasing from f to a? temp is increasing but on the same time maybe volume is decreasing so pressure can stay constant

 

[vela]

You know that f to a is an adiabat, so there's no heat transfer. The temperature is increasing, so the internal energy of the gas is increasing. Since there's no heat added to the gas, this increase in energy must be due to work done on the gas; therefore, the gas is being compressed.

 

[Andrew Mason]

how can you know pressure is increasing from f to a? temp is increasing but on the same time maybe volume is decreasing so pressure can stay constant

If S remains constant can there be heat flow? If there is no heat flow, but U increases (ie. T increases) can V remain constant (dV = 0)? If V decreases, can P decrease or stay constant if T increases? (hint: apply the ideal gas law).

If temperature is constant and entropy increases (a-b and c-d), this means that heat flows into the gas isothermally. If temperature is constant and entropy decreases (e-f) heat must flow out of the gas isothermally. If heat is flowing into or out of the gas but the temperature remains the same what must be happening to the volume? (hint: apply the first law dQ = dU + PdV where $Q \ne 0; \Delta U = 0$). What can you say about the pressure in that case?

AM

 

[shahar weiss]

Thank you very very much all of the helpers, i really appreciate it thank you

 

[Andrew Mason]

Just to finish matters for future reference:

If S remains constant there is no heat flow. This means (first law) that dU = -PdV.
Since T increases dU>0 so dV < 0. And, since PV=nRT and T increases, a decrease in V (dV<0) means that pressure must increase even more than V decreases. So in f-a P and T increase and V decreases. In d-e, T and P decrease and V increases.

If temperature is constant and entropy decreases (e-f) heat must flow out of the gas isothermally. So dU = 0 and dQ = PdV where dQ is negative. So dV must be negative - volume decreases. Conversely, if T is constant and entropy increases dQ > 0 so dV > 0 (V increases).

AM